Use the chain rule to find $\frac{d}{d x} [tan (\sqrt{x})]$ $d/dx[tan(sqrtx)]$ ?

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Answer 1 · 2018-01-11T03:38:49

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

$y = tan \sqrt{x}$ $y = tansqrtx$

Apply the chain rule and standard differential.

$\frac{d y}{d x} = {sec}^{2} \sqrt{x} \cdot \frac{d}{d x} \sqrt{x}$ $dy/dx= sec^2sqrtx *d/dx sqrtx$

$\frac{d y}{d x} = {sec}^{2} \sqrt{x} \cdot \frac{1}{2} x^{- \frac{1}{2}}$ $dy/dx = sec^2sqrtx * 1/2x^(-1/2)$

$= \frac{{sec}^{2} \sqrt{x}}{2 \sqrt{x}}$ $= (sec^2sqrtx)/(2sqrtx)$

Answer 2 · 2018-01-11T03:42:51

$\frac{d}{d x} [tan (\sqrt{x})] = \frac{{sec}^{2} (\sqrt{x})}{2 \sqrt{x}}$ $d/dx[tan(sqrtx)]=sec^2(sqrtx)/(2sqrtx)$

Apply the chain rule:

$d/dx[f(g(x))]=f'(g(x))*g'(x)$

Let $f(x)=tan(x)$ and $g(x)=sqrtx$

Thus, $f'(x)=sec^2x$ and $g'(x)=1/(2sqrtx)$

So

$d/dx[tan(sqrtx)]=sec^2(sqrtx)*1/(2sqrtx)=sec^2(sqrtx)/(2sqrtx)$

Use the chain rule to find d/dx[tan(sqrtx)]ddx[tan(√x)]d/dx[tan(sqrtx)] ?