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Q1:
Find $\frac{dy}{dx}$ in the following: $2x + 3y = \sin x$
Solution :
Step 1: Identifying the Mathematical Structure
The given equation is an implicit function of the form $f(x, y) = c$, where $2x + 3y = \sin x$. To find the derivative $\frac{dy}{dx}$, we must differentiate both sides of the equation with respect to $x$ [Per the Sum Rule and Chain Rule of Calculus].
Step 2: Differentiating Both Sides with Respect to $x$
We apply the operator $\frac{d}{dx}$ to each term independently:
$$\frac{d}{dx}(2x) + \frac{d}{dx}(3y) = \frac{d}{dx}(\sin x)$$
Step 3: Applying Derivative Rules
- For the first term, apply the Power Rule: $\frac{d}{dx}(ax^n) = anx^{n-1}$. Thus, $\frac{d}{dx}(2x) = 2(1) = 2$.
- For the second term, apply the constant multiple rule and the implicit differentiation rule: $\frac{d}{dx}(3y) = 3 \cdot \frac{dy}{dx}$.
- For the right-hand side, apply the standard trigonometric derivative: $\frac{d}{dx}(\sin x) = \cos x$.
Step 4: Algebraic Manipulation to Isolate $\frac{dy}{dx}$
Substituting the results from Step 3 into our equation:
$$2 + 3\frac{dy}{dx} = \cos x$$
Subtracting $2$ from both sides [Per the Subtraction Property of Equality]:
$$3\frac{dy}{dx} = \cos x - 2$$
Dividing by $3$ to isolate the derivative:
$$\frac{dy}{dx} = \frac{\cos x - 2}{3}$$
Visualizing the Gradient Function
Final Solution: $\frac{dy}{dx} = \frac{\cos x - 2}{3}$
More Questions from Class 12 Mathematics Continuity and Differentiability EXERCISE 5.3
- 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: $x^3 + x^2y + xy^2 + y^3 = 81$
- 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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