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