Question

x^y=logx then dy/dx at x=e is:

Answer 1

Given : $x^{y}=logx$

To find : $\frac{dy}{dx}$ at $x=e$

Solution :

• Given $x^{y}=logx$

• Taking $log$ to both sides, we get

• $ylogx=log(logx)$

• Differentiating both sides with respect to $x$, we get

• $\frac{y}{x}+\frac{dy}{dx}logx=\frac{1}{xlogx}$

• When $x=e$, we get

• $\frac{y}{e}+\frac{dy}{dx}loge=\frac{1}{eloge}$

• $\Rightarrow \frac{y}{e}+\frac{dy}{dx}=\frac{1}{e}$

• $\Rightarrow \frac{dy}{dx}=\frac{1}{e}-\frac{y}{e}$

• $\Rightarrow \boxed{\frac{dy}{dx}=\frac{1-y}{e}}$

• This is the required derivative.

Answer : $\frac{dy}{dx}=\frac{1-y}{e}$ at $x=e$