Question

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

Answer 1

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)`

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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)`

Answer 2

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)`