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Q5:
Find $\frac{dy}{dx}$ in the following: $x^2 + xy + y^2 = 100$
Solution :
Step 1: Application of the Derivative Operator
To determine $\frac{dy}{dx}$ for the implicit function $x^2 + xy + y^2 = 100$, we apply the differentiation operator $\frac{d}{dx}$ to both sides of the equation with respect to $x$:
$\frac{d}{dx}(x^2 + xy + y^2) = \frac{d}{dx}(100)$
[By the Linearity Property of Differentiation: $\frac{d}{dx}(u + v + w) = \frac{du}{dx} + \frac{dv}{dx} + \frac{dw}{dx}$]
Step 2: Term-by-Term Differentiation
We differentiate each term individually:
- $\frac{d}{dx}(x^2) = 2x$ [Using the Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$]
- $\frac{d}{dx}(xy) = x\frac{dy}{dx} + y(1)$ [Using the Product Rule: $\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}$]
- $\frac{d}{dx}(y^2) = 2y\frac{dy}{dx}$ [Using the Chain Rule: $\frac{d}{dx}(f(y)) = \frac{df}{dy} \cdot \frac{dy}{dx}$]
- $\frac{d}{dx}(100) = 0$ [Derivative of a constant is zero]
Step 3: Constructing the Algebraic Equation
Substituting the results from Step 2 into the differentiated equation:
$2x + \left( x\frac{dy}{dx} + y \right) + 2y\frac{dy}{dx} = 0$
Step 4: Isolating the $\frac{dy}{dx}$ Term
Group all terms containing $\frac{dy}{dx}$ on one side and shift remaining terms to the opposite side:
$x\frac{dy}{dx} + 2y\frac{dy}{dx} = -2x - y$
Factor out $\frac{dy}{dx}$ from the left side:
$\frac{dy}{dx}(x + 2y) = -(2x + y)$
Step 5: Solving for the Derivative
Divide both sides by the expression $(x + 2y)$ to solve for $\frac{dy}{dx}$:
$\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$
Final Solution: $\frac{dy}{dx} = -\frac{2x + y}{x + 2y}$
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: $2x + 3y = \sin y$
- Q3: Find $\frac{dy}{dx}$ in the following: $ax + by^2 = \cos y$
- Q4: Find $\frac{dy}{dx}$ in the following: $xy + y^2 = \tan x + y$
- 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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