Question

`f(x)=cos(3^x)`. Then `f'(x)=?` Do not use substitution such as `u=3^x`. Use the chain rule twice??

Answer 1

`f'(x) = -ln3 \ 3^x \ sin(3^x)`

Explanation:

The chain rule:

`d/dx f(g(x)) = f'(g(x)) \ g'(x)`

permits us to differentiate a function of a function, so for example:

`d/dx( (1+2)^10 ) = 10(1+2x)^9d/dx(1+2x)`

Where we have implicitly used the chain rule to give:

`d/dx X^n = nX^(n-1) (dX)/dx` and `X=X(x)`

In this question we are asked to differentiate using the chain rule without a substitution, ie to implicitly apply the rule. We than join (or "chain") the results together to get the final derivative:

Note that we will need the following standard derivatives:

`[A] \ d/dx cosx = -sinx`
`[B] \ d/dx a^x \ \ \ \ = lna \ a^x`

We have:

`f(x) = cos(3^x)`

So implicit application of the chain rule to differentiate wrt `a` gives:

`f'(x) = d/dx( cos(3^x) )`
`" " = -sin(3^x) d/dx (3^x)` (by [A])
`" " = -sin(3^x) \ ln3 \ 3^x` (by [B])
`" " = -ln3 \ 3^x \ sin(3^x)`