Question

Find $\frac{dy}{dx}$ of function $y=e^{x^3}+\frac{1}{2}\log x$

• A. $2.e^{x^3}{x}^2+\frac{1}{2x}$
• B. $e^{x^3}{x}^2+\frac{1}{2x}$
• C. $3.e^{x^3}{x}^2+\frac{1}{2x}$
• D. $3.e^{x^3}{x}^2+\frac{1}{x}$

Answer 1

The correct option is C $3.e^{x^3}{x}^2+\frac{1}{2x}$

Given y= $e^{x^3}+\frac{logx}{x}$

$\left(\frac{dy}{dx}\right)=\left(\frac{d{e}^{x^3}}{dx}\right)+\left(\frac{d\left(\frac{logx}{2}\right)}{dx}\right)=\left(\frac{d{e}^{x^3}}{d{x}^3}\right)\times \left(\frac{d{x}^3}{dx}\right)+\left(\frac{1}{2}\right)\left(\frac{dlogx}{dx}\right)=3x^2{e}^{x^3}+\left(\frac{1}{2x}\right)$