What is the derivative of $y = tan \sqrt{2 x + 5}$ $y=tansqrt(2x+5)$ ?

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What is the derivative of $y = tan \sqrt{2 x + 5}$ $y=tansqrt(2x+5)$ ?

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Answer 1 · 2016-01-08T11:12:00

$\frac{{sec}^{2} (\sqrt{2 x + 5})}{\sqrt{2 x + 5}}$ $sec^2(sqrt(2x+5))/sqrt(2x+5)$

Explanation:

First substitute $t = \sqrt{2 x + 5} = {(2 x + 5)}^{\frac{1}{2}}$ $t = sqrt(2x+5) = (2x+5)^(1/2)$
Then $y = tan (t)$ $y = tan(t)$
$\frac{d y}{d t} = {sec}^{2} (t)$ $dy/dt = sec^2(t)$
$\frac{d t}{d x} = \frac{1}{2} {(2 x + 5)}^{- \frac{1}{2}} \cdot 2 = \frac{1}{\sqrt{2 x + 5}}$ $dt/dx = 1/2(2x+5)^(-1/2)*2 = 1/sqrt(2x+5)$
$\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x} = \frac{{sec}^{2} (\sqrt{2 x + 5})}{\sqrt{2 x + 5}}$ $dy/dx = dy/dt*dt/dx = sec^2(sqrt(2x+5))/sqrt(2x+5)$

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What is the derivative of $y = tan \sqrt{2 x + 5}$ $y=tansqrt(2x+5)$ ?

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