Step 1: Identifying the Primary Function and Differentiation

Given the function: $y = Ae^{mx} + Be^{nx}$

We perform the first differentiation with respect to $x$ using the Chain Rule, $\frac{d}{dx}[e^{kx}] = ke^{kx}$ [By the Exponential Differentiation Rule]:

$\frac{dy}{dx} = A \cdot \frac{d}{dx}(e^{mx}) + B \cdot \frac{d}{dx}(e^{nx})$

$\frac{dy}{dx} = Am e^{mx} + Bn e^{nx}$ --- (Equation 1)

Step 2: Performing the Second Differentiation

To obtain the second derivative, differentiate Equation 1 with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(Am e^{mx} + Bn e^{nx})$

$\frac{d^2y}{dx^2} = Am \cdot \frac{d}{dx}(e^{mx}) + Bn \cdot \frac{d}{dx}(e^{nx})$

$\frac{d^2y}{dx^2} = Am^2 e^{mx} + Bn^2 e^{nx}$ --- (Equation 2)

Step 3: Substituting into the Differential Equation

We must verify if the expression $\frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny$ equals zero. Substitute the expressions derived in Step 1 and Step 2:

LHS $= (Am^2 e^{mx} + Bn^2 e^{nx}) - (m + n)(Am e^{mx} + Bn e^{nx}) + mn(Ae^{mx} + Be^{nx})$

Step 4: Algebraic Expansion and Simplification

Expand the terms within the expression:

LHS $= Am^2 e^{mx} + Bn^2 e^{nx} - [m(Am e^{mx} + Bn e^{nx}) + n(Am e^{mx} + Bn e^{nx})] + mnAe^{mx} + mnBe^{nx}$

LHS $= Am^2 e^{mx} + Bn^2 e^{nx} - (Am^2 e^{mx} + Bmne^{nx} + Amn e^{mx} + Bn^2 e^{nx}) + Amne^{mx} + Bmne^{nx}$

Distribute the negative sign across the middle bracket:

LHS $= Am^2 e^{mx} + Bn^2 e^{nx} - Am^2 e^{mx} - Bmne^{nx} - Amn e^{mx} - Bn^2 e^{nx} + Amne^{mx} + Bmne^{nx}$

Step 5: Cancellation of Terms

Group terms by their exponential components to demonstrate the cancellation:

Term Coefficient	Expression
$e^{mx}$	$Am^2 - Am^2 - Amn + Amn = 0$
$e^{nx}$	$Bn^2 - Bmn - Bn^2 + Bmn = 0$

Every term cancels out perfectly [By the Commutative and Distributive Properties of Algebra].

LHS $= 0 = \text{RHS}$

Final Solution: The identity $\frac{d^2y}{dx^2} - (m + n)\frac{dy}{dx} + mny = 0$ is proven for the given function.

Step 1: Define the Function and First Derivative

Let the given function be denoted by $f(x)$:

$f(x) = x^3 \log x$

To find the first derivative, $f'(x)$, we apply the Product Rule of differentiation, which states that for two functions $u(x)$ and $v(x)$, $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$ [Per the Product Rule of Calculus].

Let $u = x^3$ and $v = \log x$. Then:

• $\frac{du}{dx} = 3x^2$
• $\frac{dv}{dx} = \frac{1}{x}$

Applying the rule:

$f'(x) = (3x^2)(\log x) + (x^3)\left(\frac{1}{x}\right)$ $f'(x) = 3x^2 \log x + x^2$

Step 2: Determining the Second Order Derivative

The second order derivative, $f''(x)$, is defined as the derivative of the first derivative with respect to $x$:

$f''(x) = \frac{d}{dx} [3x^2 \log x + x^2]$

By the Sum Rule of Differentiation, we distribute the derivative across the terms [Per the Linearity of the Derivative Operator]:

$f''(x) = \frac{d}{dx} [3x^2 \log x] + \frac{d}{dx} [x^2]$

Step 3: Calculating Individual Components

For the first term, $3x^2 \log x$, we apply the Product Rule again:

$\frac{d}{dx} [3x^2 \log x] = \left(\frac{d}{dx} [3x^2]\right)(\log x) + (3x^2)\left(\frac{d}{dx} [\log x]\right)$ $= (6x)(\log x) + (3x^2)\left(\frac{1}{x}\right)$ $= 6x \log x + 3x$

For the second term, $x^2$, we apply the Power Rule, $\frac{d}{dx}[x^n] = nx^{n-1}$ [Per the Power Rule of Differentiation]:

$\frac{d}{dx} [x^2] = 2x$

Step 4: Final Algebraic Simplification

Combine the results from Step 3:

$f''(x) = (6x \log x + 3x) + 2x$

Combine the like terms ($3x$ and $2x$):

$f''(x) = 6x \log x + 5x$

Final Solution: $f''(x) = 6x \log x + 5x$

Step 1: Definition of the Function

Let the given function be denoted as $y$:

$y = f(x) = \log x$

The domain of this function is $x \in (0, \infty)$ [As the logarithm of non-positive real numbers is undefined in the real number system].

Step 2: Calculating the First Order Derivative

To find the first derivative, we apply the standard derivative rule for natural logarithms [By the Differentiation Power Rule and Chain Rule]:

$\frac{dy}{dx} = \frac{d}{dx}(\log x)$

$\frac{dy}{dx} = \frac{1}{x} = x^{-1}$

Step 3: Calculating the Second Order Derivative

The second order derivative is obtained by differentiating the first derivative with respect to $x$ again:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\frac{d^2y}{dx^2} = \frac{d}{dx}(x^{-1})$

Applying the Power Rule of Differentiation, where $\frac{d}{dx}(x^n) = nx^{n-1}$:

$\frac{d^2y}{dx^2} = -1 \cdot x^{-1-1}$

$\frac{d^2y}{dx^2} = -1 \cdot x^{-2}$

Step 4: Algebraic Simplification

Convert the negative exponent back into a fractional form for the final expression:

$\frac{d^2y}{dx^2} = -\frac{1}{x^2}$

Visual Representation of the Derivative Decay

Final Solution: The second order derivative of $\log x$ is $-\frac{1}{x^2}$.

Given Function

Let the given function be denoted by $y$:

$y = \sin(\log x)$

Step 1: First Order Differentiation

Differentiate $y$ with respect to $x$ using the Chain Rule of differentiation [Per the Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$].

$\frac{dy}{dx} = \frac{d}{dx}[\sin(\log x)]$

$\frac{dy}{dx} = \cos(\log x) \cdot \frac{d}{dx}(\log x)$

Since $\frac{d}{dx}(\log x) = \frac{1}{x}$ [By the standard derivative of the natural logarithm], we have:

$\frac{dy}{dx} = \frac{\cos(\log x)}{x}$

Step 2: Second Order Differentiation

To find the second order derivative, $\frac{d^2y}{dx^2}$, differentiate the expression $\frac{dy}{dx} = \frac{\cos(\log x)}{x}$ with respect to $x$ using the Quotient Rule [Per the Quotient Rule: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \cdot u' - u \cdot v'}{v^2}$].

Let $u = \cos(\log x)$ and $v = x$.

Then, $u' = \frac{d}{dx}[\cos(\log x)] = -\sin(\log x) \cdot \frac{1}{x}$

And $v' = \frac{d}{dx}[x] = 1$

Applying the Quotient Rule:

$\frac{d^2y}{dx^2} = \frac{x \cdot \left(-\sin(\log x) \cdot \frac{1}{x}\right) - \cos(\log x) \cdot (1)}{x^2}$

Step 3: Algebraic Simplification

Simplify the numerator:

$\frac{d^2y}{dx^2} = \frac{-\sin(\log x) - \cos(\log x)}{x^2}$

Factor out the negative sign to provide the standard simplified form:

$\frac{d^2y}{dx^2} = -\frac{\sin(\log x) + \cos(\log x)}{x^2}$

Final Solution: $\frac{d^2y}{dx^2} = -\frac{\sin(\log x) + \cos(\log x)}{x^2}$

Step 1: Define the Function and First Derivative

Let the given function be defined as $y = f(x) = x \cos x$.

To find the first derivative $\frac{dy}{dx}$, we apply the Product Rule of differentiation, which states that if $u$ and $v$ are functions of $x$, then $\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$ [Per the Leibniz Product Rule].

Setting $u = x$ and $v = \cos x$:

• $\frac{du}{dx} = \frac{d}{dx}(x) = 1$
• $\frac{dv}{dx} = \frac{d}{dx}(\cos x) = -\sin x$ [By standard trigonometric differentiation formula]

Applying the formula:

$\frac{dy}{dx} = x(-\sin x) + (\cos x)(1)$

$\frac{dy}{dx} = -x \sin x + \cos x$

Step 2: Define the Second Derivative

The second order derivative, denoted by $\frac{d^2y}{dx^2}$, is the derivative of the first derivative $\frac{dy}{dx}$ with respect to $x$.

$\frac{d^2y}{dx^2} = \frac{d}{dx}(-x \sin x + \cos x)$

Using the Sum Rule of differentiation [By Linearity of the Differential Operator]:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(-x \sin x) + \frac{d}{dx}(\cos x)$

Step 3: Calculating Components of the Second Derivative

For the term $\frac{d}{dx}(-x \sin x)$, we again apply the Product Rule where $u = -x$ and $v = \sin x$:

• $\frac{d}{dx}(-x) = -1$
• $\frac{d}{dx}(\sin x) = \cos x$

Calculation:

$\frac{d}{dx}(-x \sin x) = (-x)(\cos x) + (\sin x)(-1) = -x \cos x - \sin x$

Now, differentiate the remaining term $\frac{d}{dx}(\cos x) = -\sin x$.

Step 4: Algebraic Synthesis

Combine the results from Step 3:

$\frac{d^2y}{dx^2} = (-x \cos x - \sin x) + (-\sin x)$

$\frac{d^2y}{dx^2} = -x \cos x - \sin x - \sin x$

$\frac{d^2y}{dx^2} = -x \cos x - 2 \sin x$

Final Solution: $\frac{d^2y}{dx^2} = -(x \cos x + 2 \sin x)$

Step 1: First-Order Differentiation

Given the function:
$y = (\tan^{-1} x)^2$

To find the first derivative ($y_1$), we apply the Chain Rule [Per the Chain Rule: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$].

$y_1 = \frac{dy}{dx} = 2(\tan^{-1} x) \cdot \frac{d}{dx}(\tan^{-1} x)$

Recall the standard derivative of $\tan^{-1} x$: $\frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2}$.

$y_1 = 2(\tan^{-1} x) \cdot \frac{1}{1 + x^2}$

Step 2: Algebraic Rearrangement for Ease of Differentiation

To avoid the Quotient Rule in the next step, multiply both sides by $(1 + x^2)$:

$(1 + x^2) y_1 = 2 \tan^{-1} x$

Step 3: Second-Order Differentiation

Differentiate both sides with respect to $x$ using the Product Rule on the left side [Per the Product Rule: $\frac{d}{dx}(uv) = u'v + uv'$].

$\frac{d}{dx}[(1 + x^2) y_1] = \frac{d}{dx}[2 \tan^{-1} x]$

$[\frac{d}{dx}(1 + x^2)] y_1 + (1 + x^2) \frac{d}{dx}(y_1) = 2 \cdot \frac{1}{1 + x^2}$

$(2x) y_1 + (1 + x^2) y_2 = \frac{2}{1 + x^2}$

Step 4: Final Algebraic Simplification

To reach the target expression, multiply the entire equation by $(1 + x^2)$ to clear the denominator:

$(1 + x^2) \cdot [(2x) y_1 + (1 + x^2) y_2] = 2$

$(1 + x^2)(1 + x^2) y_2 + (1 + x^2)(2x) y_1 = 2$

$(x^2 + 1)^2 y_2 + 2x(x^2 + 1) y_1 = 2$

Visual Verification (Logic Mapping)

Final Solution: The derivation confirms that $(x^2 + 1)^2 y_2 + 2x(x^2 + 1)y_1 = 2$.

Step 1: Expressing the Function in Explicit Form

Given the equation: $e^y (x + 1) = 1$

To isolate $y$, we take the natural logarithm of both sides [Per the property $\ln(ab) = \ln a + \ln b$]:

$\ln(e^y) + \ln(x + 1) = \ln(1)$

$y + \ln(x + 1) = 0$ [Since $\ln(e^y) = y$ and $\ln(1) = 0$]

$y = -\ln(x + 1)$

Step 2: First Order Differentiation

Differentiate $y$ with respect to $x$ [Using the Chain Rule, $\frac{d}{dx}(\ln u) = \frac{1}{u} \cdot \frac{du}{dx}$]:

$\frac{dy}{dx} = -\frac{1}{x+1} \cdot \frac{d}{dx}(x+1)$

$\frac{dy}{dx} = -\frac{1}{x+1}$

Step 3: Second Order Differentiation

Differentiate the first derivative $\frac{dy}{dx}$ with respect to $x$ again:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \left[ -(x+1)^{-1} \right]$

Using the power rule $\frac{d}{dx}(u^n) = nu^{n-1} \cdot \frac{du}{dx}$:

$\frac{d^2y}{dx^2} = -(-1)(x+1)^{-2} \cdot (1)$

$\frac{d^2y}{dx^2} = \frac{1}{(x+1)^2}$

Step 4: Verification of the Identity

Observe the square of the first derivative calculated in Step 2:

$\left( \frac{dy}{dx} \right)^2 = \left( -\frac{1}{x+1} \right)^2$

$\left( \frac{dy}{dx} \right)^2 = \frac{1}{(x+1)^2}$

Comparing the results from Step 3 and Step 4, we see that:

$\frac{d^2y}{dx^2} = \frac{1}{(x+1)^2} = \left( \frac{dy}{dx} \right)^2$

Graphical Representation of the Relationship

Final Solution: Since $\frac{d^2y}{dx^2} = \frac{1}{(x+1)^2}$ and $\left(\frac{dy}{dx}\right)^2 = \left(-\frac{1}{x+1}\right)^2 = \frac{1}{(x+1)^2}$, it is proven that $\frac{d^2y}{dx^2} = \left(\frac{dy}{dx}\right)^2$.

Step 1: Define the Function

Let the given function be denoted by $y$:

$y = f(x) = x^{20}$

Step 2: First Order Differentiation

To find the first derivative $\frac{dy}{dx}$, we apply the Power Rule of differentiation, which states that if $f(x) = x^n$, then $f'(x) = nx^{n-1}$ [By Power Rule of Calculus].

$\frac{dy}{dx} = \frac{d}{dx}(x^{20})$

$\frac{dy}{dx} = 20x^{20-1}$

$\frac{dy}{dx} = 20x^{19}$

Step 3: Second Order Differentiation

The second order derivative, denoted as $\frac{d^2y}{dx^2}$, is obtained by differentiating the first derivative with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right)$

$\frac{d^2y}{dx^2} = \frac{d}{dx}(20x^{19})$

Applying the Constant Multiple Rule [$\frac{d}{dx}(cf(x)) = c \frac{d}{dx}f(x)$]:

$\frac{d^2y}{dx^2} = 20 \cdot \frac{d}{dx}(x^{19})$

Applying the Power Rule again:

$\frac{d^2y}{dx^2} = 20 \cdot (19x^{19-1})$

$\frac{d^2y}{dx^2} = 20 \cdot 19 \cdot x^{18}$

Step 4: Algebraic Simplification

Perform the scalar multiplication of the constants:

$20 \times 19 = 380$

Therefore:

$\frac{d^2y}{dx^2} = 380x^{18}$

Visual Representation of Power Function Growth

Final Solution: The second order derivative of $x^{20}$ is $380x^{18}$.

Step 1: Establishing the Initial Function

The given function is:
$y = 500e^{7x} + 600e^{-7x}$ --- (Equation 1)

Step 2: First Order Differentiation

Differentiating Equation 1 with respect to $x$ using the Chain Rule $\frac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot \frac{du}{dx}$ [Per the Chain Rule of Differentiation]:
$\frac{dy}{dx} = 500 \cdot \frac{d}{dx}(e^{7x}) + 600 \cdot \frac{d}{dx}(e^{-7x})$
$\frac{dy}{dx} = 500(e^{7x} \cdot 7) + 600(e^{-7x} \cdot -7)$
$\frac{dy}{dx} = 3500e^{7x} - 4200e^{-7x}$ --- (Equation 2)

Step 3: Second Order Differentiation

Differentiating Equation 2 with respect to $x$ to obtain the second-order derivative:
$\frac{d^2y}{dx^2} = \frac{d}{dx}(3500e^{7x} - 4200e^{-7x})$
$\frac{d^2y}{dx^2} = 3500 \cdot \frac{d}{dx}(e^{7x}) - 4200 \cdot \frac{d}{dx}(e^{-7x})$
$\frac{d^2y}{dx^2} = 3500(e^{7x} \cdot 7) - 4200(e^{-7x} \cdot -7)$

Step 4: Algebraic Simplification

Multiplying the terms:
$\frac{d^2y}{dx^2} = 24500e^{7x} + 29400e^{-7x}$

Step 5: Factorization and Verification

To demonstrate that $\frac{d^2y}{dx^2} = 49y$, we factor out $49$ from the expression obtained in Step 4:
$\frac{d^2y}{dx^2} = 49(500e^{7x} + 600e^{-7x})$

Comparing the expression within the parentheses to Equation 1, we observe:
$500e^{7x} + 600e^{-7x} = y$

Therefore:
$\frac{d^2y}{dx^2} = 49y$

Visual Representation of the Derivative Relationship

Final Solution: Since $\frac{d^2y}{dx^2} = 24500e^{7x} + 29400e^{-7x}$ and $y = 500e^{7x} + 600e^{-7x}$, it is proven that $\frac{d^2y}{dx^2} = 49y$.

Step 1: Define the Function

Let the given function be $y$ such that:

$y = \log(\log x)$

where $x > 1$ for the function to be defined in the real domain [Per the domain constraints of the logarithmic function $\log(u)$ where $u > 0$].

Step 2: First-Order Differentiation

To find the first derivative $\frac{dy}{dx}$, we apply the Chain Rule. Recall that $\frac{d}{dx}[\log(u)] = \frac{1}{u} \cdot \frac{du}{dx}$ [By the Chain Rule of Differentiation].

$\frac{dy}{dx} = \frac{d}{dx}[\log(\log x)]$

$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{d}{dx}[\log x]$

Since $\frac{d}{dx}[\log x] = \frac{1}{x}$, we have:

$\frac{dy}{dx} = \frac{1}{\log x} \cdot \frac{1}{x} = \frac{1}{x \log x} = (x \log x)^{-1}$

Step 3: Second-Order Differentiation

Now, we find $\frac{d^2y}{dx^2}$ by differentiating $\frac{dy}{dx}$ with respect to $x$ using the Quotient Rule or the Power Rule with Chain Rule. We shall use the Quotient Rule: $\frac{d}{dx}\left[\frac{u}{v}\right] = \frac{v \cdot u' - u \cdot v'}{v^2}$ [By the Quotient Rule of Differentiation].

Let $u = 1$ and $v = x \log x$.

Then, $\frac{du}{dx} = 0$.

To find $\frac{dv}{dx}$, we use the Product Rule: $\frac{d}{dx}[x \log x] = x \cdot \frac{d}{dx}[\log x] + (\log x) \cdot \frac{d}{dx}[x]$ [By the Product Rule of Differentiation].

$\frac{dv}{dx} = x \cdot \frac{1}{x} + \log x \cdot 1 = 1 + \log x$.

Step 4: Algebraic Substitution and Simplification

Applying the Quotient Rule formula:

$\frac{d^2y}{dx^2} = \frac{(x \log x)(0) - (1)(1 + \log x)}{(x \log x)^2}$

$\frac{d^2y}{dx^2} = \frac{-(1 + \log x)}{(x \log x)^2}$

Visual Representation of Function Growth

Final Solution: $\frac{d^2y}{dx^2} = -\frac{1 + \log x}{(x \log x)^2}$

Given Variables & Initial Setup

The provided function is: $y = 3 \cos(\log x) + 4 \sin(\log x)$

We are required to prove that $x^2 y_2 + x y_1 + y = 0$, where $y_1$ denotes the first derivative $\frac{dy}{dx}$ and $y_2$ denotes the second derivative $\frac{d^2y}{dx^2}$.

Step 1: Differentiation with respect to $x$

Applying the Chain Rule [$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$] and the derivative of the natural logarithm [$\frac{d}{dx}(\log x) = \frac{1}{x}$]:

$y_1 = \frac{dy}{dx} = 3 \left( -\sin(\log x) \cdot \frac{1}{x} \right) + 4 \left( \cos(\log x) \cdot \frac{1}{x} \right)$

$y_1 = \frac{-3 \sin(\log x) + 4 \cos(\log x)}{x}$

Step 2: Rearranging the First Derivative

To simplify the subsequent differentiation, multiply both sides by $x$:

$x y_1 = -3 \sin(\log x) + 4 \cos(\log x)$

Step 3: Differentiation with respect to $x$ (Second Order)

Differentiating both sides with respect to $x$ using the Product Rule on the left side [$\frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx}$]:

$\frac{d}{dx}(x y_1) = \frac{d}{dx}(-3 \sin(\log x) + 4 \cos(\log x))$

$(x \cdot y_2 + y_1 \cdot 1) = -3 \cos(\log x) \cdot \frac{1}{x} - 4 \sin(\log x) \cdot \frac{1}{x}$

Step 4: Algebraic Simplification

Multiply the entire equation by $x$ to eliminate the denominator:

$x(x y_2 + y_1) = -3 \cos(\log x) - 4 \sin(\log x)$

$x^2 y_2 + x y_1 = -(3 \cos(\log x) + 4 \sin(\log x))$

Step 5: Substitution and Final Proof

Observe that the expression in the parenthesis is the original function $y$:

$x^2 y_2 + x y_1 = -(y)$

Adding $y$ to both sides yields:

$x^2 y_2 + x y_1 + y = 0$

Visual Representation of Function Structure

Final Solution: Through successive differentiation and substitution, we have successfully shown that $x^2 y_2 + x y_1 + y = 0$.

Step 1: Identifying the Function and First Order Derivative

Given the function $f(x) = \tan^{-1} x$. To find the second order derivative, we must first determine the first order derivative with respect to $x$.

Applying the standard derivative formula for inverse trigonometric functions [By the Rule of Derivatives of Inverse Functions]:

$$\frac{d}{dx} (\tan^{-1} x) = \frac{1}{1 + x^2}$$

Let $y = \tan^{-1} x$. Therefore, the first derivative is:

$$y' = \frac{dy}{dx} = (1 + x^2)^{-1}$$

Step 2: Applying the Chain Rule for Second Order Differentiation

To find the second order derivative $\frac{d^2y}{dx^2}$, we differentiate $y'$ with respect to $x$ [By the Chain Rule of Differentiation]:

$$\frac{d^2y}{dx^2} = \frac{d}{dx} \left[ (1 + x^2)^{-1} \right]$

Using the Power Rule $\frac{d}{dx}(u^n) = n u^{n-1} \cdot \frac{du}{dx}$, where $u = 1 + x^2$ and $n = -1$:

$$\frac{d^2y}{dx^2} = -1 \cdot (1 + x^2)^{-1 - 1} \cdot \frac{d}{dx}(1 + x^2)$$

Step 3: Evaluating the Differential Component

Calculating the derivative of the inner function $(1 + x^2)$ with respect to $x$ [By the Sum Rule and Power Rule of Differentiation]:

$$\frac{d}{dx}(1 + x^2) = 0 + 2x = 2x$$

Substituting this result back into our expression for the second derivative:

$$\frac{d^2y}{dx^2} = -1 \cdot (1 + x^2)^{-2} \cdot (2x)$$

Step 4: Simplifying the Final Expression

Combining the terms into a standard rational algebraic form:

$$\frac{d^2y}{dx^2} = \frac{-2x}{(1 + x^2)^2}$$

Final Solution: The second order derivative of $\tan^{-1} x$ is $\frac{-2x}{(1 + x^2)^2}$.

Step 1: Define the Function and First Derivative

Let the given function be $y = f(x) = e^{6x} \cos 3x$.

To find the first derivative $\frac{dy}{dx}$, we apply the Product Rule of differentiation: $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$. [Per the Product Rule]

Let $u = e^{6x}$ and $v = \cos 3x$. Then:

• $\frac{du}{dx} = 6e^{6x}$ [By the Chain Rule: $\frac{d}{dx}e^{ax} = ae^{ax}$]
• $\frac{dv}{dx} = -3\sin 3x$ [By the Chain Rule: $\frac{d}{dx}\cos(ax) = -a\sin(ax)$]

Substituting these into the product rule formula:

$\frac{dy}{dx} = e^{6x}(-3\sin 3x) + (\cos 3x)(6e^{6x})$

$\frac{dy}{dx} = e^{6x} (6\cos 3x - 3\sin 3x)$

Step 2: Define the Second Derivative

To find the second order derivative $\frac{d^2y}{dx^2}$, we differentiate the first derivative with respect to $x$ again:

$\frac{d^2y}{dx^2} = \frac{d}{dx} [e^{6x} (6\cos 3x - 3\sin 3x)]$

Applying the Product Rule again, let $u = e^{6x}$ and $v = (6\cos 3x - 3\sin 3x)$.

Then $\frac{du}{dx} = 6e^{6x}$ and $\frac{dv}{dx} = \frac{d}{dx}[6\cos 3x - 3\sin 3x] = 6(-3\sin 3x) - 3(3\cos 3x) = -18\sin 3x - 9\cos 3x$.

Step 3: Algebraic Expansion and Simplification

Using the Product Rule expression $\frac{d^2y}{dx^2} = u \frac{dv}{dx} + v \frac{du}{dx}$:

$\frac{d^2y}{dx^2} = e^{6x}(-18\sin 3x - 9\cos 3x) + (6\cos 3x - 3\sin 3x)(6e^{6x})$

Factor out $e^{6x}$:

$\frac{d^2y}{dx^2} = e^{6x} [-18\sin 3x - 9\cos 3x + 6(6\cos 3x - 3\sin 3x)]$

$\frac{d^2y}{dx^2} = e^{6x} [-18\sin 3x - 9\cos 3x + 36\cos 3x - 18\sin 3x]$

Step 4: Combining Like Terms

Combine the $\sin 3x$ and $\cos 3x$ terms:

$\sin 3x \text{ components: } -18 - 18 = -36$

$\cos 3x \text{ components: } -9 + 36 = 27$

Resulting in:

$\frac{d^2y}{dx^2} = e^{6x} [27\cos 3x - 36\sin 3x]$

Factor out the common constant $9$:

$\frac{d^2y}{dx^2} = 9e^{6x} (3\cos 3x - 4\sin 3x)$

Final Solution: \frac{d^2y}{dx^2} = 9e^{6x}(3\cos 3x - 4\sin 3x)

Step 1: First Order Differentiation

Given the function: $y = \cos^{-1} x$

To find the first derivative, differentiate both sides with respect to $x$ using the standard derivative formula for inverse trigonometric functions [Derivative of $\cos^{-1}x = -\frac{1}{\sqrt{1-x^2}}$]:

$\frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}$

We can rewrite this as: $\frac{dy}{dx} = -(1-x^2)^{-1/2}$

Step 2: Second Order Differentiation

Differentiate $\frac{dy}{dx}$ with respect to $x$ again to obtain $\frac{d^2y}{dx^2}$. Apply the Power Rule $\frac{d}{dx}[u^n] = nu^{n-1} \cdot \frac{du}{dx}$ and the Chain Rule:

$\frac{d^2y}{dx^2} = \frac{d}{dx} [-(1-x^2)^{-1/2}]$

$\frac{d^2y}{dx^2} = - \left( -\frac{1}{2} \right) (1-x^2)^{-3/2} \cdot \frac{d}{dx}(1-x^2)$ [By Chain Rule]

$\frac{d^2y}{dx^2} = \frac{1}{2} (1-x^2)^{-3/2} \cdot (-2x)$

$\frac{d^2y}{dx^2} = -x(1-x^2)^{-3/2} = -\frac{x}{(1-x^2)^{3/2}}$

Step 3: Conversion to Terms of $y$

From the original equation $y = \cos^{-1} x$, we know that $x = \cos y$.

Substitute $x = \cos y$ into the expression for the second derivative:

$\frac{d^2y}{dx^2} = -\frac{\cos y}{(1-\cos^2 y)^{3/2}}$

Applying the Pythagorean identity $[\sin^2 y + \cos^2 y = 1 \implies 1-\cos^2 y = \sin^2 y]$:

$\frac{d^2y}{dx^2} = -\frac{\cos y}{(\sin^2 y)^{3/2}}$

$\frac{d^2y}{dx^2} = -\frac{\cos y}{(\sin y)^3}$

Step 4: Algebraic Simplification

Decompose the denominator into $\sin y \cdot \sin^2 y$:

$\frac{d^2y}{dx^2} = -\left( \frac{\cos y}{\sin y} \right) \cdot \left( \frac{1}{\sin^2 y} \right)$

Using trigonometric identities $[\cot y = \frac{\cos y}{\sin y}$ and $\csc^2 y = \frac{1}{\sin^2 y}]$:

$\frac{d^2y}{dx^2} = -\cot y \csc^2 y$

Final Solution: \frac{d^2y}{dx^2} = -\cot y \csc^2 y

Step 1: Establishing the Initial Function

The given function is defined as:

$y = 5 \cos x - 3 \sin x$ — (Eq. 1)

Step 2: Calculating the First Derivative

Differentiate $y$ with respect to $x$ using the basic differentiation rules for trigonometric functions, specifically $\frac{d}{dx}(\cos x) = -\sin x$ and $\frac{d}{dx}(\sin x) = \cos x$ [By the Sum/Difference Rule of Differentiation]:

$\frac{dy}{dx} = \frac{d}{dx}(5 \cos x) - \frac{d}{dx}(3 \sin x)$

$\frac{dy}{dx} = 5(-\sin x) - 3(\cos x)$

$\frac{dy}{dx} = -5 \sin x - 3 \cos x$ — (Eq. 2)

Step 3: Calculating the Second Derivative

Differentiate the first derivative $\frac{dy}{dx}$ with respect to $x$ to obtain the second-order derivative $\frac{d^2y}{dx^2}$:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(-5 \sin x - 3 \cos x)$

Applying the linearity of the derivative operator:

$\frac{d^2y}{dx^2} = -5 \frac{d}{dx}(\sin x) - 3 \frac{d}{dx}(\cos x)$

$\frac{d^2y}{dx^2} = -5(\cos x) - 3(-\sin x)$

$\frac{d^2y}{dx^2} = -5 \cos x + 3 \sin x$ — (Eq. 3)

Step 4: Algebraic Verification of the Identity

We are required to prove that $\frac{d^2y}{dx^2} + y = 0$. Substituting Eq. 3 and Eq. 1 into the expression:

$\text{LHS} = \frac{d^2y}{dx^2} + y$

$\text{LHS} = (-5 \cos x + 3 \sin x) + (5 \cos x - 3 \sin x)$

Grouping the terms by trigonometric functions:

$\text{LHS} = (-5 \cos x + 5 \cos x) + (3 \sin x - 3 \sin x)$

$\text{LHS} = 0 + 0 = 0$

$\text{LHS} = \text{RHS}$

Visualization of the Function

Final Solution: Since the substitution of the second derivative and the original function results in zero, the identity $\frac{d^2y}{dx^2} + y = 0$ is proven.

Step 1: Establishing the Initial Function

Let the given function be denoted by $y$:

$y = e^x \sin(5x)$

Step 2: First Order Differentiation

To find the first derivative $\frac{dy}{dx}$, we apply the Product Rule for differentiation, which states that for two differentiable functions $u(x)$ and $v(x)$, $\frac{d}{dx}[u(x)v(x)] = u(x)v'(x) + v(x)u'(x)$ [Per the Product Rule of Calculus].

Let $u = e^x$ and $v = \sin(5x)$.

$\frac{du}{dx} = e^x$

$\frac{dv}{dx} = \frac{d}{dx}[\sin(5x)] = \cos(5x) \cdot \frac{d}{dx}(5x) = 5\cos(5x)$ [Per the Chain Rule of Calculus]

Substituting these into the product rule formula:

$\frac{dy}{dx} = e^x(5\cos(5x)) + \sin(5x)(e^x)$

Factor out the common term $e^x$:

$\frac{dy}{dx} = e^x [5\cos(5x) + \sin(5x)]$

Step 3: Second Order Differentiation

To find the second derivative $\frac{d^2y}{dx^2}$, differentiate $\frac{dy}{dx}$ with respect to $x$ using the Product Rule again.

Let $u = e^x$ and $v = [5\cos(5x) + \sin(5x)]$.

Then, $u' = e^x$.

$v' = \frac{d}{dx}[5\cos(5x) + \sin(5x)] = 5(-\sin(5x) \cdot 5) + (\cos(5x) \cdot 5)$

$v' = -25\sin(5x) + 5\cos(5x)$

Applying the Product Rule: $\frac{d^2y}{dx^2} = u \cdot v' + v \cdot u'$

$\frac{d^2y}{dx^2} = e^x [-25\sin(5x) + 5\cos(5x)] + [5\cos(5x) + \sin(5x)] e^x$

Step 4: Algebraic Simplification

Distribute and collect like terms:

$\frac{d^2y}{dx^2} = e^x [-25\sin(5x) + 5\cos(5x) + 5\cos(5x) + \sin(5x)]$

Combine the $\sin(5x)$ and $\cos(5x)$ terms:

$\frac{d^2y}{dx^2} = e^x [10\cos(5x) - 24\sin(5x)]$

Final Solution: $\frac{d^2y}{dx^2} = e^x [10\cos(5x) - 24\sin(5x)]$

Step 1: Define the Function

Let the given function be represented as $f(x)$. We define the function as follows:

$f(x) = x^2 + 3x + 2$

Step 2: First Order Differentiation

To find the first order derivative ($f'(x)$ or $\frac{dy}{dx}$), we apply the Power Rule of differentiation, which states that $\frac{d}{dx}(x^n) = nx^{n-1}$ [By the Power Rule].

$\frac{d}{dx}(x^2 + 3x + 2) = \frac{d}{dx}(x^2) + \frac{d}{dx}(3x) + \frac{d}{dx}(2)$

$\frac{dy}{dx} = 2x^{(2-1)} + 3(1)x^{(1-1)} + 0$ [Since the derivative of a constant is 0]

$\frac{dy}{dx} = 2x + 3$

Step 3: Second Order Differentiation

To find the second order derivative ($\frac{d^2y}{dx^2}$), we differentiate the first order derivative with respect to $x$ again.

$\frac{d^2y}{dx^2} = \frac{d}{dx}(2x + 3)$

Applying the linearity of the derivative operator:

$\frac{d^2y}{dx^2} = \frac{d}{dx}(2x) + \frac{d}{dx}(3)$

$\frac{d^2y}{dx^2} = 2(1) + 0$

$\frac{d^2y}{dx^2} = 2$

Visual Representation of the Function and its Derivatives

Final Solution: The second order derivative of $x^2 + 3x + 2$ is $2$.