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

##### 1 Answer

#### Answer:

# 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

# 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) #