Calculus Help!

<p>My teacher gave us a worksheet, and no one has been able to find out the last question. It'll probably prove to be pretty easy for most of you, but our class is behind...</p>

<p>Anyway, here's a rundown of what was asked and what I have in order to answer the last question:</p>

<p>Find these derivatives:<br>
h(x)=(2x+1)^3 I got: h'(x)=6(2x+1)^2
h(x)=(x^2+5)^2 I got: h'(x)=4x(x^2+5)
h(x)=sin(2x) I got: h'(x)=2(-sinx^2+cosx^2)</p>

<p>Let each of the functions above represent a composition of f o g.</p>

<p>Define functions f and g for #1-3 above. (I dunno if I did these correctly)
1. f(x)=6x^2 g(x)=2x+1
2. f(x)=4x g(x)=x^3+5x
3. f(x)=2x g(x)=-sinx^2+cosx^2 (I think this one is wrong...)</p>

<p><strong><em>!!!Give a formula for finding h'(x) in terms of f and g.!!!</em></strong>
I don't know how to get this.</p>

<p>We are just going into the section regarding the chain rule, so we arent allowed to use the chain rule or anything beyond that. She mentioned looking at the pattern between the derivatives and the original functions in the first three questions.</p>

<p>Can anyone help?</p>

<p>err, for h(x) = sin(2x), h'(x) = 2cos(2x)</p>

<p>ok, h(x) = f(g(x)), right? but I think you solved it so h'(x) = f(g(x)). That's probably why you're getting confused.</p>

<p>i think the last one is 2cos2x?
because if you take the derivative of sin2x as cos2x you still have to multiply it by the derivative of the 2x which is 2...</p>

<p>His answer, h'=2(-sinx^2+cosx^2), is correct because cos2x=cosx^2-sinx^2.</p>

<p>nice catch, averagemathgeek. It's been so long since I've done those stupid trig identities that I had completely forgotten that. =`></p>

<p>So my answers for the second part should be compositions for the original functions and not the derivatives I found?</p>

<p>If not, then the last part surely cant be as easy as f(g(x))=h'(x)...?</p>

<p>The problem is, you (or your worksheet) is too general... it says "the functions above." Now, I would assume that is referring to the original functions, and not the derivatives, but I wasn't in your class so I can't be sure.</p>

<p>I assume it means the original functions (I copied directly from the worksheet, and she didnt say otherwise), because the fifth part was supposedly tricky. It seems the fifth part is cake when I use compositions of the derivatives...</p>

<p>Can anyone find the answer to the last question when using the composition of the original functions?</p>