How do you differentiate $f (x) = csc (\sqrt{2 x})$ $f(x)=csc(sqrt(2x))$ using the chain rule?

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Answer 1 · 2016-01-02T20:12:29

$f'(x)=(-csc(sqrt(2x))cot(sqrt(2x)))/(sqrt(2x))$

According to the chain rule,

$d/dx(cscu)=-u'cscucotu$

Thus,

$f'(x)=d/dx(csc(sqrt(2x)))=-csc(sqrt(2x))cot(sqrt(2x))*d/dx(sqrt(2x))$

To find $d/dx(sqrt(2x))$ , use the chain rule again:

$d/dx(sqrtu)=1/(2sqrtu)*u'$

So,

$d/dx(sqrt(2x))=1/(2sqrt(2x))*2=1/(sqrt(2x)$

Plug this back in to find $f'(x)$ :

$f'(x)=(-csc(sqrt(2x))cot(sqrt(2x)))/(sqrt(2x))$

How do you differentiate f(x)=csc(sqrt(2x)) f(x)=csc(√2x)f(x)=csc(sqrt(2x)) using the chain rule?