How do you differentiate $f (x) = \sqrt{sin (e^{4 x})}$ $f(x)=sqrtsin(e^(4x))$ using the chain rule.?

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How do you differentiate $f (x) = \sqrt{sin (e^{4 x})}$ $f(x)=sqrtsin(e^(4x))$ using the chain rule.?

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Answer 1 · 2016-02-09T19:20:41

[2e^(4x)cos(e(^4x))] / [sqrt sin(e^(4x)]

Explanation:

Applying chain rule
df(u)/dx= df/du .du/dx
let sin e^(4x) =u
d/du √u . d/dx (sin(e^(4x)))
we have,
d/du √u=1/(2√u)
and
d/dx (sin(e^(4x)))

Applying chain rule,
df(u)/dx= df/du .du/dx
let $e^{4 x} = u$ $e^{4x}=u$
$\frac{d}{d u}$ $\frac{d}{du}$ (sin(u) $\frac{d}{d x}$ $\frac{d}{dx}$

$\frac{d}{d u}$ $\frac{d}{du}$ sin(u) = cos(u)
$\frac{d}{d x} (e^{4 x})$ $\frac{d}{dx} (e^{4x})$
Applying chain rule,
df(u)/dx= df/du .du/dx
let 4x=u
solving it we get,
$e^{4 x}$ $e^{4x}$ 4
now, cos(u) $e^{4 x} 4$ $e^{4x}4$
u= $e^{4 x}$ $e^{4x}$
= cos( $e^{4 x}$ $e^{4x}$ ) $e^{4 x} 4$ $e^{4x}4$

Finally,
$\frac{1}{2 \sqrt{u}}$ $\frac{1}{2\sqrt{u}}$ cos( $e^{4 x}$ $e^{4x}$ ) $e^{4 x} 4$ $e^{4x}4$
Substitute u= sin( $(e^{4 x})$ $(e^{4x})$
= $\frac{1}{2 \sqrt{sin (e^{4 x})}}$ $\frac{1}{2\sqrt{\sin (e^{4x})}}$ cos( $(e^{4 x}) e^{4 x} 4$ $(e^{4x})e^{4x}4$
simplifying we get,
$\frac{2 e^{4 x} cos (e^{4 x})}{\sqrt{sin (e^{4 x})}}$ $\frac{2e^{4x}\cos(e^{4x})}{\sqrt{\sin (e^{4x})}}$

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How do you differentiate $f (x) = \sqrt{sin (e^{4 x})}$ $f(x)=sqrtsin(e^(4x))$ using the chain rule.?

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How do you differentiate f(x)=√sin(e4x)f(x)=sqrtsin(e^(4x)) using the chain rule.?

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How do you differentiate $f (x) = \sqrt{sin (e^{4 x})}$ $f(x)=sqrtsin(e^(4x))$ using the chain rule.?