Question

How do you differentiate `f(x) = ln(sqrt(arcsin(e^(3x)) )` using the chain rule?

Answer 1

`f'(x)=(3e^(3x))/(2sqrt(1-e^6)arcsin(e^(3x)))`

Explanation:

The first step we should take is to simplify the function by using logarithm rules to remove the square root from the function.

Since the square root is really a `1"/"2` power, and `ln(a^b)=bln(a)`, this function can be rewritten as

`f(x)=1/2ln(arcsin(e^(3x)))`

From here, the first issue is the natural logarithm. According to the chain rule, we know that `d/dx[ln(u)]=1/u*u'`, and we have `u=arcsin(e^(3x))`.

`f'(x)=1/2(1/arcsin(e^(3x)))d/dx[arcsin(e^(3x))]`

Now, to differentiate the `arcsin` function, we will again use the chain rule, in that `d/dx[arcsin(u)]=1/(sqrt(1-x^2))*u'`, and this time `u=e^(3x)`.

`f'(x)=1/(2arcsin(e^(3x)))(1/sqrt(1-(e^(3x))^2))d/dx[e^(3x)]`

Before moving on to find the derivative of `e^(3x)`, simplify a little...

`f'(x)=1/(2sqrt(1-e^6)arcsin(e^(3x)))*d/dx[e^(3x)]`

This final bout of differentiation also requires the chain rule, but differentiating things to the `e` power is e-asy: `d/dx[e^u]=e^u*u'`, where `u=3x`.

`f'(x)=1/(2sqrt(1-e^6)arcsin(e^(3x)))(e^(3x)d/dx[3x])`

Since `d/dx[3x]=3`, this all simplifies to be

`f'(x)=(3e^(3x))/(2sqrt(1-e^6)arcsin(e^(3x)))`