Question

Find $\frac{dy}{dx}$ for $y{e}^{x^2}+\tan x+\log \left(\sin x\right)=0$

• A. $-\left(\frac{y{e}^{x^2}.2x-{\sec }^{2}x+\cot x}{e^{x^2}}\right)$
• B. $-\left(\frac{y{e}^{x^2}.2x+{\sec }^{2}x+\cot x}{e^{x^2}}\right)$
• C. $\left(\frac{y{e}^{x^2}.2x+{\sec }^{2}x+\cot x}{e^{x^2}}\right)$
• D. None of these

Answer 1

Answer: (B)

$-\left(\frac{y{e}^{x^2}.2x+{\sec }^{2}x+\cot x}{e^{x^2}}\right)$

Given $y{e}^{x^2}+\tan x+\log \left(\sin x\right)=0$

Using product rule and chain rule:

$\Rightarrow y{e}^{x^2}2x+\frac{dy}{dx}{e}^{x^2}+{\sec }^{2}x+\frac{\cos x}{\sin x}=0$

$\Rightarrow y{e}^{x^2}2x+\frac{dy}{dx}{e}^{x^2}+{\sec }^{2}x+\cot x=0$

$\Rightarrow \frac{dy}{dx}=-\left(\frac{y{e}^{x^2}2x+{\sec }^{2}x+\cot x}{e^{x^2}}\right)$