How do you differentiate #f(x)=e^tan(sqrtx) # using the chain rule?

1 Answer
Apr 3, 2016

Answer:

#f'(x)=(e^tan(sqrtx)sec^2(sqrtx))/(2sqrtx)#

Explanation:

We have the chain rule working on multiple levels. The first level is that we have an exponential function in the form #e^u#, where #u=tan(sqrtx)#.

The chain rule states that

#d/dx(e^u)=e^u*u'#

So, we see that

#f'(x)=e^tan(sqrtx)*d/dx(tan(sqrtx))#

Now, we have to deal with differentiating the tangent function in the form #tan(u)#, where #u=sqrtx#.

Again, through the chain rule, we see that

#d/dx(tan(u))=sec^2(u)*u'#

This gives us

#d/dx(tan(sqrtx))=sec^2(sqrtx)*d/dx(sqrtx)#

To find #d/dx(sqrtx)#, recall that #sqrtx=x^(1/2)#, so #d/dx(x^(1/2))=1/2x^(-1/2)=1/(2sqrtx)#.

Combining everything we've found, we see that

#f'(x)=e^tan(sqrtx)*sec^2(sqrtx)*1/(2sqrtx)#

#f'(x)=(e^tan(sqrtx)sec^2(sqrtx))/(2sqrtx)#