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

Aug 12, 2017

$f ' \left(x\right) = - \ln 3 \setminus {3}^{x} \setminus \sin \left({3}^{x}\right)$

#### Explanation:

The chain rule:

$\frac{d}{\mathrm{dx}} f \left(g \left(x\right)\right) = f ' \left(g \left(x\right)\right) \setminus g ' \left(x\right)$

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

$\frac{d}{\mathrm{dx}} \left({\left(1 + 2\right)}^{10}\right) = 10 {\left(1 + 2 x\right)}^{9} \frac{d}{\mathrm{dx}} \left(1 + 2 x\right)$

Where we have implicitly used the chain rule to give:

$\frac{d}{\mathrm{dx}} {X}^{n} = n {X}^{n - 1} \frac{\mathrm{dX}}{\mathrm{dx}}$ and $X = X \left(x\right)$

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:

$\left[A\right] \setminus \frac{d}{\mathrm{dx}} \cos x = - \sin x$
$\left[B\right] \setminus \frac{d}{\mathrm{dx}} {a}^{x} \setminus \setminus \setminus \setminus = \ln a \setminus {a}^{x}$

We have:

$f \left(x\right) = \cos \left({3}^{x}\right)$

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

$f ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(\cos \left({3}^{x}\right)\right)$
$\text{ } = - \sin \left({3}^{x}\right) \frac{d}{\mathrm{dx}} \left({3}^{x}\right)$ (by [A])
$\text{ } = - \sin \left({3}^{x}\right) \setminus \ln 3 \setminus {3}^{x}$ (by [B])
$\text{ } = - \ln 3 \setminus {3}^{x} \setminus \sin \left({3}^{x}\right)$