How do you differentiate $f (x) = \sqrt{csc (2 x - 4)}$ $f(x)=sqrtcsc(2x -4)$ using the chain rule?

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Answer 1 · 2018-02-09T01:51:15

$f'(x)=-cos(2x-4)csc(2x-4)^(3/2)$

rewrite the function like this
$csc(2x-4)^(1/2)$

use chain rule by first using power rule on the $1/2$ exponent,
then on the csc function, and finally on the function inside the csc

$cancel(1/2)csc(2x-4)^(-1/2)*(-csc(2x-4)cot(2x-4))cancel(2)$

then

$csc(2x-4)^(-1/2)=1/sqrt(csc(2x-4))$

multiply

$(-csc(2x-4)cot(2x-4))/(sqrt(csc(2x-4))$

convert everything into sin and cos
$((-1/sin(2x-4))*(cos(2x-4)/sin(2x-4)))/csc(2x-4)^(1/2)$

simplify

$((-cos(2x-4))/(sin^2(2x-4)))/(1/sin(2x-4)^(1/2))$

flip and multiply

$(-cos(2x-4)sin(2x-4)^(1/2))/sin^2(2x-4)$

subtract sin exponents

$-cos(2x-4)*sin(2x-4)^(-3/2)$

more simplification

$-cos(2x-4)*(1/csc(2x-4)^(-3/2))$

$-cos(2x-4)csc(2x-4)^(3/2)$

How do you differentiate f(x)=sqrtcsc(2x -4) f(x)=√csc(2x−4)f(x)=sqrtcsc(2x -4) using the chain rule?