How do you integrate #f(x)=x(x^7+15)^3# using the product rule?

1 Answer
Jan 13, 2018

Answer:

#f'(x)=(x^7+15)^3+21x^7(x^7+15)^2#

Explanation:

I'm assuming that you meant differentiate, because we cannot integrate using the product rule.

#f(x)=x(x^7+15)^3#

Use the product rule, which states that

#d/dx(uv)=u'v+uv'#

Let #u=x#, #v=(x^7+15)^3#

We now need to differentiate #v=(x^7+15)^3#, so we are going to use the chain rule, which states that

#dy/dx=dy/(du)*(du)/dx#

Since we already have a #u# for the product rule, and we are trying to differentiate #v#, let's rewrite the rule as

#(dv)/dx=(dv)/(da)*(da)/(dx)#

Let #a=x^7+15#, so #(da)/dx=7x^6#, #v=a^3#, #(dv)/(da)=3a^2#

Combining, we get #(dv)/dx=7x^6*3a^2#

#(dv)/dx=21x^6a^2#

Putting back #a=x^7+15#, we get

#(dv)/dx=21x^6(x^7+15)^2#

#v'=21x^6(x^7+15)^2#

Now that we have differentiated #v#, let's put it in the product rule

#d/dx(f(x))=1*(x^7+15)^3+x(21x^6(x^7+15)^2)#

#f'(x)=(x^7+15)^3+21x^7(x^7+15)^2#