What is the derivative of the exponential function y = e^(4tansqrtx)?

Apr 14, 2015

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{4 \tan \sqrt{x}} {\sec}^{2} \sqrt{x}}{\sqrt{x}}$

Solution

$y = {e}^{4 \tan \sqrt{x}}$

Differentiating both sides with respect to 'x'

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{d}{\mathrm{dx}} \left({e}^{4 \tan \sqrt{x}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{4 \tan \sqrt{x}} \frac{d}{\mathrm{dx}} \left(4 \tan \sqrt{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{4 \tan \sqrt{x}} .4 {\sec}^{2} \sqrt{x} . \frac{d}{\mathrm{dx}} \left(\sqrt{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 4 {e}^{4 \tan \sqrt{x}} {\sec}^{2} \sqrt{x} \left(\frac{1}{2} {x}^{\frac{1}{2} - 1}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{4}{2} {e}^{4 \tan \sqrt{x}} {\sec}^{2} \sqrt{x} \left({x}^{\frac{1 - 2}{2}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 {e}^{4 \tan \sqrt{x}} {\sec}^{2} \sqrt{x} \left({x}^{\frac{- 1}{2}}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{4 \tan \sqrt{x}} {\sec}^{2} \sqrt{x}}{{x}^{\frac{1}{2}}}$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 {e}^{4 \tan \sqrt{x}} {\sec}^{2} \sqrt{x}}{\sqrt{x}}$