# Thread: Find the derivative of the function below.

1. ## Find the derivative of the function below.

h(y)=(b/(a+y^4))^4
find h'(y)

I was using the chain rule

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

and I got...

4(b/(a+y^4))((-4by^3)/(a+y^4)^2)

could someone tell me the answer, or if you have time, tell me step by step what to do?

2. Originally Posted by Latszer
h(y)=(b/(a+y^4))^4
find h'(y)

I was using the chain rule

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

and I got...

4(b/(a+y^4))((-4by^3)/(a+y^4)^2)

could someone tell me the answer, or if you have time, tell me step by step what to do?
You're so close!!

It should be $\displaystyle 4\left(\frac{b}{a+y^4}\right)^{{\color{red}3}}\lef t(-\frac{4by^3}{\left(a+y^4\right)^2}\right)$

3. Originally Posted by Latszer
h(y)=(b/(a+y^4))^4
find h'(y)

I was using the chain rule

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

and I got...

4(b/(a+y^4))((-4by^3)/(a+y^4)^2)

could someone tell me the answer, or if you have time, tell me step by step what to do?
$\displaystyle h'\left( y \right) = - \frac{{4{b^4}}}{{{{\left( {a + {y^4}} \right)}^5}}} \cdot {\left( {a + {y^4}} \right)^\prime } = - \frac{{4{b^4}}}{{{{\left( {a + {y^4}} \right)}^5}}} \cdot 4{y^3} = - \frac{{16{b^4}{y^3}}}{{{{\left( {a + {y^4}} \right)}^5}}}.$

4. Yeah, made an error and didn't put it to the third, I guess I had it right. Thanks you two.