How do you differentiate #g(x) = sqrtxsqrt(1-e^(2x))# using the product rule?

1 Answer
Nov 23, 2017

Answer:

#(1-e^(2x)-2xe^(2x))/(2sqrtxsqrt(1-e^(2x)))#

Explanation:

The product rule states that:
#d/dx(f(x)g(x))=d/dx(f(x))g(x)+f(x)d/dx(g(x))#

Plugging in your above expression, we get:
#d/dx(sqrt(x)sqrt(1-e^(2x)))=d/dx(sqrt(x))sqrt(1-e^(2x))+sqrt(x)d/dx(sqrt(1-e^(2x)))#

The #sqrtx# derivative is quite easy if you notice that #sqrtx=x^(1/2)# and use the power rule. However, the other derivative is a bit more complex. We need to use the chain rule to evaluate it:
#d/dx(sqrt(1-e^(2x)))#

We can let #u=1-e^(2x)#, and rewrite the expression like this:
#d/(du)(sqrt(u))d/dx(1-e^(2x))=1/(2sqrtu)*d/dx(1-e^(2x))#

Now we once again need to use the chain rule. If we let #z=2x#, we get:
#1/(2sqrtu)*d/(dz)(1-e^z)d/dx(2x)=-1/(2sqrtu)2e^z#

Resubstituting, we get:
#-1/(2sqrt(1-e^(2x)))*2e^(2x)=-e^(2x)/sqrt(1-e^(2x))#

Now that we have evaluated the difficult part in our product rule expression, we can put it back in and see that the answer becomes this:
#1/(2sqrtx)sqrt(1-e^(2x))-sqrtxe^(2x)/sqrt(1-e^(2x))=sqrt(1-e^(2x))/(2sqrtx)-(sqrtxe^(2x))/sqrt(1-e^(2x))#

If we multiply by the opposite denominator, we can get the two fractions to have the same denominator:
#(sqrt(1-e^(2x))sqrt(1-e^(2x)))/(2sqrtxsqrt(1-e^(2x)))-(2sqrtxsqrtxe^(2x))/(2sqrtxsqrt(1-e^(2x)))=(1-e^(2x)-2xe^(2x))/(2sqrtxsqrt(1-e^(2x)))#