How do you differentiate 5cos (2x)+ tan(3x+4)?

2 Answers
Jun 2, 2015

Let f(u(x)) = 5cos(2x) + tan(3x+4).

For this, you would have to remember the following two derivatives:

(d/(dx))[cosu] = -sinu(du(x))/(dx)
(d/(dx))[tanu] = sec^2u(du(x))/(dx)

since as you can see, (df(u(x)))/(dx) = (df(u(x)))/cancel(du(x))*cancel(du(x))/(dx)

where u(x) = 2x or u(x) = 3x + 4, in this case, for example. That just explains why the chain rule works how it works.

So, we get:

(d/(dx))[5cos(2x) + tan(3x+4)]

= 5*-sin(2x)*2 + sec^2(3x+4)*3

= -10sin(2x) + 3sec^2(3x+4)


Other ways you might see the chain rule:

In "prime" notation:

[f(u(x))]' = f'(u(x))*u'(x)

or in "function composition" notation:

[f(x)@u(x)]' = [f'(x)@u(x)]*u'(x)

Jun 2, 2015

I would use the Chain Rule in both terms as:
y'=-5sin(2x)*2+1/(cos^2(3x+4))*3=
=-10sin(2x)+3/(cos^2(3x+4))

Where the Chain Rule allows you to derive y=f(g(x)) as:
y'=f'(g(x))*g'(x)