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Q7:
Find $\frac{dy}{dx}$ in the following: $\sin^2 y + \cos xy = \kappa$
Solution :
Step 1: Application of Implicit Differentiation
Given the equation: $\sin^2 y + \cos(xy) = \kappa$, where $\kappa$ is a constant. We differentiate both sides with respect to $x$ using the chain rule and the sum rule [Per the Rule of Differentiation of Implicit Functions].
$\frac{d}{dx}(\sin^2 y) + \frac{d}{dx}(\cos(xy)) = \frac{d}{dx}(\kappa)$
Step 2: Differentiating Term by Term
For the first term $\frac{d}{dx}(\sin^2 y)$: Applying the power rule followed by the chain rule: $\frac{d}{dx}(\sin^2 y) = 2 \sin y \cdot \frac{d}{dx}(\sin y) = 2 \sin y \cdot \cos y \cdot \frac{dy}{dx}$ Using the double angle identity $\sin(2\theta) = 2\sin\theta\cos\theta$, we simplify: $= \sin(2y) \frac{dy}{dx}$
For the second term $\frac{d}{dx}(\cos(xy))$: Applying the chain rule and product rule: $\frac{d}{dx}(\cos(xy)) = -\sin(xy) \cdot \frac{d}{dx}(xy)$ Using the product rule for $\frac{d}{dx}(xy) = x \frac{dy}{dx} + y(1)$: $= -\sin(xy) \cdot (x \frac{dy}{dx} + y)$
For the constant term: $\frac{d}{dx}(\kappa) = 0$ [Per the Constant Rule of Differentiation]
Step 3: Constructing the Linear Equation for $\frac{dy}{dx}$
Substituting the derivatives back into the original expression: $\sin(2y) \frac{dy}{dx} - \sin(xy)(x \frac{dy}{dx} + y) = 0$
Distributing $-\sin(xy)$: $\sin(2y) \frac{dy}{dx} - x \sin(xy) \frac{dy}{dx} - y \sin(xy) = 0$
Step 4: Isolating $\frac{dy}{dx}$
Group all terms containing $\frac{dy}{dx}$ on one side: $\frac{dy}{dx} (\sin(2y) - x \sin(xy)) = y \sin(xy)$
Solving for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}$
Visual Representation of the Differential Geometry
Final Solution: \frac{dy}{dx} = \frac{y \sin(xy)}{\sin(2y) - x \sin(xy)}
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: $xy + y^2 = \tan x + y$
- Q5: Find $\frac{dy}{dx}$ in the following: $x^2 + xy + y^2 = 100$
- Q6: Find $\frac{dy}{dx}$ in the following: $x^3 + x^2y + xy^2 + y^3 = 81$
- 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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