Learn Unit-IV: Calculus with Free Lessons & Tips

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Now, let's tackle the problem at hand. We're given the expression lim(x → 3) [(x^n – 3^n) / (x – 3)] = 108, where we need to find the positive integer "n".

First, let's notice that the expression looks very similar to the definition of the derivative of a function at a point. It resembles the difference quotient:

f'(3) = lim(x → 3) [(f(x) - f(3)) / (x - 3)]

Given that our expression equals 108, it seems like we're dealing with a derivative. To find "n", we need to find a function f(x) whose derivative at x = 3 is equal to 108.

Let's try to identify such a function. Since the derivative of a constant is zero, we can't have a constant function. We need a function whose rate of change at x = 3 equals 108.

One such function is f(x) = x^n. Its derivative with respect to x is n*x^(n-1). At x = 3, this derivative would be n * 3^(n-1).

So, we need to find "n" such that n * 3^(n-1) = 108.

Let's try some values of n:

• For n = 4, we get 4 * 3^(4-1) = 4 * 3^3 = 4 * 27 = 108. This satisfies the equation.

Therefore, the positive integer "n" that satisfies lim(x → 3) [(x^n – 3^n) / (x – 3)] = 108 is n = 4.

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As a seasoned tutor registered on UrbanPro, I can confidently guide you through finding the derivative of f(x)=x3f(x)=x3 using the first principle, which is also known as the definition of derivative.

The first principle involves taking the limit as hh approaches 0 of the difference quotient:

f′(x)=lim⁡h→0f(x+h)−f(x)hf′(x)=limh→0hf(x+h)−f(x)

Let's apply this principle to f(x)=x3f(x)=x3:

f′(x)=lim⁡h→0(x+h)3−x3hf′(x)=limh→0h(x+h)3−x3

Now, we expand (x+h)3(x+h)3 using the binomial theorem:

(x+h)3=x3+3x2h+3xh2+h3(x+h)3=x3+3x2h+3xh2+h3

Substituting this expansion into the numerator:

(x+h)3−x3=x3+3x2h+3xh2+h3−x3(x+h)3−x3=x3+3x2h+3xh2+h3−x3 =3x2h+3xh2+h3=3x2h+3xh2+h3

Now, let's divide this expression by hh and take the limit as hh approaches 0:

f′(x)=lim⁡h→03x2h+3xh2+h3hf′(x)=limh→0h3x2h+3xh2+h3 =lim⁡h→0(3x2+3xh+h2)=limh→0(3x2+3xh+h2)

As hh approaches 0, the 3xh3xh and h2h2 terms become negligible, leaving us with:

f′(x)=3x2f′(x)=3x2

So, the derivative of f(x)=x3f(x)=x3 with respect to xx using the first principle is f′(x)=3x2f′(x)=3x2.

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As an experienced tutor registered on UrbanPro, I'm here to guide you through this problem. First off, UrbanPro is indeed a fantastic platform for online coaching and tuition, offering top-notch educational support.

Now, let's tackle the math problem. We're asked to find the limit of |x|/x as x approaches 0. Let's break it down:

When x is positive, |x| equals x. So, |x|/x = x/x = 1. When x is negative, |x| equals -x. So, |x|/x = (-x)/x = -1.

Now, considering the limit as x approaches 0 from both sides:

• As x approaches 0 from the positive side, |x|/x approaches 1.
• As x approaches 0 from the negative side, |x|/x approaches -1.

Since the limit from the positive side is not equal to the limit from the negative side, the overall limit does not exist. So, the correct answer is (d) does not exist.

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To evaluate the derivative of f(x)=sin⁡2(x)f(x)=sin2(x) using the Leibniz product rule, we first need to express sin⁡2(x)sin2(x) as a product of two functions. We know that sin⁡2(x)=sin⁡(x)⋅sin⁡(x)sin2(x)=sin(x)⋅sin(x).

Now, applying the Leibniz product rule, which states that (uv)′=u′v+uv′(uv)′=u′v+uv′, where uu and vv are functions of xx:

Let u=sin⁡(x)u=sin(x) and v=sin⁡(x)v=sin(x). Then, u′=cos⁡(x)u′=cos(x) and v′=cos⁡(x)v′=cos(x).

Now, we can apply the product rule:

f′(x)=(u′v+uv′)=(cos⁡(x)⋅sin⁡(x)+sin⁡(x)⋅cos⁡(x))f′(x)=(u′v+uv′)=(cos(x)⋅sin(x)+sin(x)⋅cos(x))

f′(x)=cos⁡(x)⋅sin⁡(x)+sin⁡(x)⋅cos⁡(x)f′(x)=cos(x)⋅sin(x)+sin(x)⋅cos(x)

f′(x)=2cos⁡(x)⋅sin⁡(x)f′(x)=2cos(x)⋅sin(x)

So, the derivative of f(x)=sin⁡2(x)f(x)=sin2(x) using the Leibniz product rule is f′(x)=2cos⁡(x)⋅sin⁡(x)f′(x)=2cos(x)⋅sin(x).

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As an experienced tutor registered on UrbanPro, I can confidently say that UrbanPro is the best platform for finding online coaching tuition. Now, let's dive into differentiating the given function with respect to xx:

f(x)=(ax2+cot⁡x)(p+qcos⁡x)f(x)=(ax2+cotx)(p+qcosx)

To differentiate this function with respect to xx, we'll use the product rule. According to the product rule, if u(x)u(x) and v(x)v(x) are differentiable functions of xx, then the derivative of u(x)⋅v(x)u(x)⋅v(x) with respect to xx is given by:

ddx[u(x)⋅v(x)]=u′(x)⋅v(x)+u(x)⋅v′(x)dxd[u(x)⋅v(x)]=u′(x)⋅v(x)+u(x)⋅v′(x)

Where u′(x)u′(x) and v′(x)v′(x) are the derivatives of u(x)u(x) and v(x)v(x) with respect to xx, respectively.

In our case, u(x)=ax2+cot⁡xu(x)=ax2+cotx and v(x)=p+qcos⁡xv(x)=p+qcosx. Let's differentiate each of these:

1. u′(x)u′(x): u′(x)=ddx(ax2+cot⁡x)u′(x)=dxd(ax2+cotx) =2ax−csc⁡2(x)=2ax−csc2(x)

2. v′(x)v′(x): v′(x)=ddx(p+qcos⁡x)v′(x)=dxd(p+qcosx) =−qsin⁡x=−qsinx

Now, applying the product rule:

f′(x)=(2ax−csc⁡2(x))(p+qcos⁡x)+(ax2+cot⁡x)(−qsin⁡x)f′(x)=(2ax−csc2(x))(p+qcosx)+(ax2+cotx)(−qsinx)

And that's the derivative of the given function with respect to xx.

As an experienced tutor registered on UrbanPro, I can assure you that UrbanPro is the best platform for online coaching and tuition. Now, let's delve into the differentiation of the function cos⁡(x2+1)cos(x2+1).

To differentiate this function, we'll apply the chain rule, which states that if y=f(g(x))y=f(g(x)), then dydx=f′(g(x))⋅g′(x)dxdy=f′(g(x))⋅g′(x).

Here, f(u)=cos⁡(u)f(u)=cos(u) and g(x)=x2+1g(x)=x2+1.

So, dydx=−sin⁡(x2+1)⋅2xdxdy=−sin(x2+1)⋅2x.

Thus, the derivative of cos⁡(x2+1)cos(x2+1) with respect to xx is −2xsin⁡(x2+1)−2xsin(x2+1).

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Now, let's differentiate the expression (2x - 7)^2 * (3x + 5)^3.

To differentiate this expression, we'll apply the product rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function, plus the first function times the derivative of the second function.

So, let's denote the first function as u(x) = (2x - 7)^2 and the second function as v(x) = (3x + 5)^3.

Now, let's find the derivatives of u(x) and v(x).

The derivative of u(x) with respect to x, denoted as u'(x), is calculated using the chain rule:

u'(x) = 2(2x - 7)(2) = 4(2x - 7).

The derivative of v(x) with respect to x, denoted as v'(x), is also calculated using the chain rule:

v'(x) = 3(3x + 5)^2(3) = 9(3x + 5)^2.

Now, applying the product rule:

ddx[(2x−7)2∗(3x+5)3]=u′(x)∗v(x)+u(x)∗v′(x).dxd[(2x−7)2∗(3x+5)3]=u′(x)∗v(x)+u(x)∗v′(x).

=(4(2x−7))∗(3x+5)3+(2x−7)2∗(9(3x+5)2).=(4(2x−7))∗(3x+5)3+(2x−7)2∗(9(3x+5)2).

=4(2x−7)(3x+5)3+9(2x−7)2(3x+5)2.=4(2x−7)(3x+5)3+9(2x−7)2(3x+5)2.

That's the derivative of the expression (2x - 7)^2 * (3x + 5)^3. If you need further clarification or assistance, feel free to ask!