How do you differentiate # f(x)=x*(9 + 2 x)*sqrt(x^2-3)# using the product rule?

1 Answer
Oct 27, 2015

Answer:

#f'(x)=(3(2x^3 +6x^2 -4x-9))/(sqrt(x^2-3))#

Explanation:

We have a few options. We can leave the function as it is and use the product rule twice, or you can simplify by multiplying the #x# through #9+2x# and using the product rule once. Since the second is easier, lets go with that one.

The product rule tells us that if we have a function #f(x)#, where;

#f(x)=g(x)*h(x)#

Then;

#f'(x)=g(x)*h'(x) + g'(x)*h(x)#

If we simplify the function as mentioned above, by multiplying the #x# through, we get;

#f(x)=(9x+2x^2)sqrt(x^2-3)#

Now we can split our function into;

#g(x)=9x+2x^2#
#h(x)=sqrt(x^2-3)#

If we use the power rule to find #g'(x)# we get;

#g'(x)=9+4x#

Finding #h'(x)# is a little bit trickier since we have to use the chain rule . If we have a function #f(x)# where;

#f(x)=g(h(x))#

Then;

#f'(x)=g'(h(x))*h'(x)#

We can rewrite #h(x)# to make things a little clearer.

#h(x)=(x^2-3)^(1/2)#

Let #u(x)=x^2-3#.

#h(u)=u^(1/2)#

Applying the power rule;

#h'(u)=1/2u^((1/2-1))u'=1/cancel2(x^2-3)^(-1/2)(cancel2x)=x/sqrt(x^2-3)#

Now that we have #h'# and #g'# we can put everything together.

#f'(x)=g(x)*h'(x) + g'(x)*h(x)#

#=(9x+2x^2)(x)/sqrt(x^2-3) +(9+4x)sqrt(x^2-3)#

All that's left is to simplify.

#=(9x^2 + 2x^3)/sqrt(x^2-3)+(9+4x)sqrt(x^2-3)#

#=(9x^2 + 2x^3)/sqrt(x^2-3)+(9+4x)sqrt(x^2-3)*(sqrt(x^2-3))/(sqrt(x^2-3))#

#=(9x^2 + 2x^3)/sqrt(x^2-3)+((9+4x)(x^2-3))/(sqrt(x^2-3))#

#=((9x^2 + 2x^3)+9x^2+4x^3-27-12x)/(sqrt(x^2-3))#

#=(6x^3 +18x^2 - 12x-27)/(sqrt(x^2-3))#

#=(3(2x^3 +6x^2 -4x-9))/(sqrt(x^2-3))#

*Note If you choose to use the product rule twice;

#g(x)=x(9+2x) = j(x)*k(x)#

Where;

#j(x)=x#
#k(x)=9+2x#

So;

#g'(x)=j(x)k'(x)+j'(x)k(x)#

And;

#f'(x)=g(x)*h'(x) + (j(x)k'(x)+j'(x)k(x))*h(x)#

You end up with the same answer, it just takes more work. This method would be good if you couldn't easily simplify the original function.