What is the derivative of the exponential function $y = e^{4 tan \sqrt{x}}$ $y = e^(4tansqrtx)$ ?

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Answer 1 · 2015-04-14T16:04:23

$\frac{d y}{d x} = \frac{2 e^{4 tan \sqrt{x}} {sec}^{2} \sqrt{x}}{\sqrt{x}}$ $dy/dx=(2e^(4tansqrtx)sec^2sqrtx)/sqrtx$

Solution

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

Differentiating both sides with respect to 'x'

$\frac{d y}{d x} = \frac{d}{d x} (e^{4 tan \sqrt{x}})$ $dy/dx=d/dx(e^(4tansqrtx))$

$\frac{d y}{d x} = e^{4 tan \sqrt{x}} \frac{d}{d x} (4 tan \sqrt{x})$ $dy/dx=e^(4tansqrtx)d/dx(4tansqrtx)$

$\frac{d y}{d x} = e^{4 tan \sqrt{x}} .4 {sec}^{2} \sqrt{x} . \frac{d}{d x} (\sqrt{x})$ $dy/dx=e^(4tansqrtx).4sec^2sqrtx.d/dx(sqrtx)$

$\frac{d y}{d x} = 4 e^{4 tan \sqrt{x}} {sec}^{2} \sqrt{x} (\frac{1}{2} x^{\frac{1}{2} - 1})$ $dy/dx=4e^(4tansqrtx)sec^2sqrtx(1/2x^(1/2-1))$

$\frac{d y}{d x} = \frac{4}{2} e^{4 tan \sqrt{x}} {sec}^{2} \sqrt{x} (x^{\frac{1 - 2}{2}})$ $dy/dx=4/2e^(4tansqrtx)sec^2sqrtx(x^((1-2)/2))$

$\frac{d y}{d x} = 2 e^{4 tan \sqrt{x}} {sec}^{2} \sqrt{x} (x^{\frac{- 1}{2}})$ $dy/dx=2e^(4tansqrtx)sec^2sqrtx(x^((-1)/2))$

$\frac{d y}{d x} = \frac{2 e^{4 tan \sqrt{x}} {sec}^{2} \sqrt{x}}{x^{\frac{1}{2}}}$ $dy/dx=(2e^(4tansqrtx)sec^2sqrtx)/(x^((1)/2))$

$\frac{d y}{d x} = \frac{2 e^{4 tan \sqrt{x}} {sec}^{2} \sqrt{x}}{\sqrt{x}}$ $dy/dx=(2e^(4tansqrtx)sec^2sqrtx)/sqrtx$

What is the derivative of the exponential function y = e^(4tansqrtx)y=e4tan√xy = e^(4tansqrtx)?