Re: Rules of derivatives.

You don't *need* chain rule to find the derivative of g(x) since the x is to the first power and has no constant in front of it, you could just use power rule to get g'(x) to be 3(x+5)^2. You then HAVE to use quotient rule.

Re: Rules of derivatives.

I'm not sure I'm getting that yet. Is it really possible to just use the power rule. I was under the impression that (x+5)^3 would have to be expanded by the binomial theorem or alternatively the chain rule would have to be applied. The term I question has two terms parenthesized to the third power. I was under the impression that the power rule works on individual terms only.

I think what your saying is that since x is to the first power and has no constant in front of it that the power rule can still apply but otherwise it couldn't.

Re: Rules of derivatives.

Quote:

Originally Posted by

**sepoto** I'm not sure I'm getting that yet. Is it really possible to just use the power rule. I was under the impression that (x+5)^3 would have to be expanded by the binomial theorem or alternatively the chain rule would have to be applied. The term I question has two terms parenthesized to the third power. I was under the impression that the power rule works on individual terms only.

I think what your saying is that since x is to the first power and has no constant in front of it that the power rule can still apply but otherwise it couldn't.

Use the chain rule...it's easier this way. We have that g(x):

The 1 at the end is the derivative of x + 5.

-Dan

Re: Rules of derivatives.

Mikewezyk's point is that, in using the chain rule, you don't **have** to write out every step. Taking u= x+ 5, . By the chain rule, the derivative of is . Mikewezyk's point is that the derivative of x+ 5 is just 1. I would not agree that this is "not using the chain rule" but rather using the chain rule without writing everything out.

More generally, if g(x)= f(ax+ b) then taking u= ax+ b, g(x)= f(u) so that, by the chain rule, g'(x)= f'(u) du/dx= a f'(u). But with a little practice, you can do that "in your head". The derivative of the **inear** term ax+ b is a so that g'(x)= af'(ax+ b).