How do you differentiate # g(x) = 2 csc ^2x + 5 cos x #?

Question

1852 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-14T18:52:25

#g'(x)=-4csc^2(x)cot(x)-5sin(x)#

It helps to know some trig derivatives:

$tutorial.math.lamar.edu$

We're also going to need product rule since there's a #csc^2x#. Rewrite #g(x)#:

#g(x)=2csc(x)csc(x)+5cos(x)#

Now take the derivative with product rule:

#g'(x)=-2csc^2(x)cot(x)-2csc^2(x)cot(x)-5sin(x)#

Simplify:

#g'(x)=-4csc^2(x)cot(x)-5sin(x)#