Thread: chain rule

attchement

Originally Posted by Mathmania

attchement

1.) g(x) = (3 + sqrt(x))^4

g'(x) = 4*(3 + sqrt(x))^3 * {derivative of inside here}

Remember that sqrt(x) is x^(1/2).

= 4*(3 + sqrt(x))^3 * [1/(2*sqrt(x))]

Thus, g'(x) = [2*(3 + sqrt(x))^3]/[sqrt(x)]

2.) g(x) = [6*x]/[(x^3 - 5)^(1/4)]

Lets change it a little. We could do the quotient rule, but that's boring and more tedious in my opinion.

g(x) = [6*x]*[(x^3 - 5)^(-1/4)]

For the derivative, you have to do the product rule AND the chain rule.

Derivative of the 1st term, times the second, plus the derivative of the second term times the 1st.

6*[(x^3 - 5)^(-1/4)] + [(-1/4)*(x^3 - 5)^(-5/4)*3*x^2]*(6*x)

Simplify that and you're done.

It simplifies to:

[3*x^3 - 60]/[2*(x^3 - 5)^(5/4)]

3.) f(x) = [(3*x^2 + 5)^4]/[(2-3*x^4)^3];

I'll use quotient rule this time to show some variety and hopefully help you solve some of the other questions on your own.

The quotient rule states:

[f(x)/g(x)]' = [f'(x)*g(x) - f(x)*g'(x)]/[(g(x))^2]

In other words, simply put, the derivative of the top times the bottom minus the top times the derivative of the bottom.

For your problem:

f'(x) = [[4*(3*x^2 + 5)^3*6*x]*[(2-3*x^4)^3] - [(3*x^2 + 5)^4]*[3*(2-3*x^4)^2*-12*x^3]]/[(2-3*x^4)^3]^2

It says to leave it in unsimplified form. So there you have it. Remember when you take the derivative of a composition of terms to a power, you have to take the power, place it in front, multiply by the original, reduce the power, and then remember to go back and take the derivative of the inside.

4.) Your turn: by now you should be able to do #4 very quickly.

Do it yourself.

I will give you what the answer will be. See if your answer matches up.

h'(1) = 7.408.

EDIT: It asks for h'(1) to two decimal places. Thus, 7.41.