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Q3:
Find $\frac{dy}{dx}$ in the following: $ax + by^2 = \cos y$
Solution :
Given Equation & Differentiation Strategy
The given equation is:
$ax + by^2 = \cos y$
To find $\frac{dy}{dx}$, we must perform implicit differentiation with respect to $x$ across the entire equation [Per the Chain Rule of Calculus].
Step 1: Differentiating Both Sides with Respect to $x$
Applying the operator $\frac{d}{dx}$ to both sides of the equation:
$\frac{d}{dx}(ax + by^2) = \frac{d}{dx}(\cos y)$
Step 2: Applying Sum and Constant Multiple Rules
Distributing the derivative across the terms and factoring out constants:
$a \cdot \frac{d}{dx}(x) + b \cdot \frac{d}{dx}(y^2) = \frac{d}{dx}(\cos y)$
[By the Linearity Property of Differentiation]
Step 3: Executing the Chain Rule
Differentiate each term individually, keeping in mind that $y$ is a function of $x$ ($y=f(x)$):
- $\frac{d}{dx}(ax) = a(1) = a$
- $b \cdot \frac{d}{dx}(y^2) = b \cdot (2y \cdot \frac{dy}{dx}) = 2by \frac{dy}{dx}$ [By Power Rule and Chain Rule]
- $\frac{d}{dx}(\cos y) = -\sin y \cdot \frac{dy}{dx}$ [By Derivative of Trigonometric Functions and Chain Rule]
Substituting these back into the equation:
$a + 2by \frac{dy}{dx} = -\sin y \frac{dy}{dx}$
Step 4: Isolating the $\frac{dy}{dx}$ terms
Group all terms containing $\frac{dy}{dx}$ on one side of the equation:
$a = -\sin y \frac{dy}{dx} - 2by \frac{dy}{dx}$
$a = -\frac{dy}{dx} (\sin y + 2by)$
Step 5: Solving for $\frac{dy}{dx}$
Divide both sides by the coefficient of $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{a}{ -(\sin y + 2by)}$
$\frac{dy}{dx} = -\frac{a}{2by + \sin y}$
Geometric Representation of Implicit Differentiation
Final Solution: $\frac{dy}{dx} = -\frac{a}{2by + \sin y}$
More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.3
- Q1: Find $\frac{dy}{dx}$ in the following: $2x + 3y = \sin x$
- Q2: Find $\frac{dy}{dx}$ in the following: $2x + 3y = \sin y$
- Q4: Find $\frac{dy}{dx}$ in the following: $xy + y^2 = \tan x + y$
- Q5: Find $\frac{dy}{dx}$ in the following: $x^2 + xy + y^2 = 100$
- Q6: Find $\frac{dy}{dx}$ in the following: $x^3 + x^2y + xy^2 + y^3 = 81$
- Q7: Find $\frac{dy}{dx}$ in the following: $\sin^2 y + \cos xy = \kappa$
- Q8: Find $\frac{dy}{dx}$ in the following: $\sin^2 x + \cos^2 y = 1$
- Q9: Find $\frac{dy}{dx}$ in the following: $y = \sin^{-1} \left(\frac{2x}{1+x^2}\right)$
- Q10: Find $\frac{dy}{dx}$ in the following: $y = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right), -\frac{1}{\sqrt{3}} < x < \frac{1}{\sqrt{3}}$
- Q11: Find $\frac{dy}{dx}$ in the following: $y = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$
- Q12: Find $\frac{dy}{dx}$ in the following: $y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right), 0 < x < 1$
- Q13: Find $\frac{dy}{dx}$ in the following: $y = \cos^{-1}\left(\frac{2x}{1+x^2}\right), -1 < x < 1$
- Q14: Find $\frac{dy}{dx}$ in the following: $y = \sin^{-1}(2x\sqrt{1-x^2}), -\frac{1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$
- Q15: Find $\frac{dy}{dx}$ in the following: $y = \sec^{-1}\left(\frac{1}{2x^2-1}\right), 0 < x < \frac{1}{\sqrt{2}}$
CBSE Solutions for Class 12 Mathematics Continuity and Differentiability
- Continuity and Differentiability Class 12 EXERCISE 5.1
- Continuity and Differentiability Class 12 EXERCISE 5.2
- Continuity and Differentiability Class 12 EXERCISE 5.4
- Continuity and Differentiability Class 12 EXERCISE 5.5
- Continuity and Differentiability Class 12 EXERCISE 5.6
- Continuity and Differentiability Class 12 EXERCISE 5.7
Chapters in CBSE - Class 12 Mathematics
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