# Math Help - chain rule problem, with 2 constants!

1. ## chain rule problem, with 2 constants!

$\frac{d}{dw}{[a\cos^{2}\pi w +b\sin^{2}\pi w ]}$ (a,b constants)

The constant "a" or "b" independent from cos/sin here?..

Is this is correct?

$\frac{d}{dw}{[(a\cos\pi w)^{2} +(b\sin\pi w)^{2} ]}$

OR this one is the correct one?

$\frac{d}{dw}{[a(\cos\pi w)^{2} +b(\sin\pi w)^{2} ]}$

2. ## Re: chain rule problem, with 2 constants!

The second one is equivalent to the original statement.

3. ## Re: chain rule problem, with 2 constants!

then we have to use the product rule right?

4. ## Re: chain rule problem, with 2 constants!

No, you have a sum, not a product. Differentiating term by term, I would use the power and chain rules.

5. ## Re: chain rule problem, with 2 constants!

Originally Posted by MarkFL2
No, you have a sum, not a product. Differentiating term by term, I would use the power and chain rules.
Right sum but isn't there a product rule too here?
$\frac{d}{dw}{[a(\cos\pi w)^{2} ]}$

$a \and \(\cos\pi w)^{2}$

6. ## Re: chain rule problem, with 2 constants!

If $a$ and $b$ are constants, while you could technically use the product rule, I would simply use the rule:

$\frac{d}{dx}(k\cdot f(u))=k\cdot\frac{d}{dx}(f(u))$ where $k$ is an arbitrary constant.

You will get the same result using the product rule since the derivative of a constant is zero.

7. ## Re: chain rule problem, with 2 constants!

I solved the problem but my result is differing from the text book..

I got

${-2a\cos\pi w \sin\pi w + 2b \sin\pi w \cos\pi w}$

The text book has

${-2\pi a\cos\pi w \sin\pi w + 2\pi b \sin\pi w \cos\pi w}$

an extra $\pi$ between 2a and 2b.

8. ## Re: chain rule problem, with 2 constants!

You have only partially applied the chain rule. You must also apply it to the arguments of the trig. functions. This is where the $\pi$ comes from that you are missing.

9. ## Re: chain rule problem, with 2 constants!

this is what I did
$\frac{d}{dw}(\cos \pi w)$

then

$-\sin \pi w$

I should have to applied product rule here? or maybe rewrite $\frac{d}{dw}(\cos \pi w)$ to $\frac{d}{dw}(\cos \pi) \frac{d}{dw}(\cos w)$

is that is my mistake?

10. ## Re: chain rule problem, with 2 constants!

You need in the case from your previous post:

$\frac{d}{dw}(\cos(\pi w))=-\sin(\pi w)\cdot\frac{d}{dw}(\pi w)=-\pi\sin(\pi w)$

11. ## Re: chain rule problem, with 2 constants!

Originally Posted by MarkFL2
You need in the case from your previous post:

$\frac{d}{dw}(\cos(\pi w))=-\sin(\pi w)\cdot\frac{d}{dw}(\pi w)=-\pi\sin(\pi w)$
I find it confusing to decide how deep I should use the change rule. What Should I know.. how do I decide..

if let say some constant 5 for example was here instead of w.. so it would have looked like $\pi 5$

what could been the scenario?

12. ## Re: chain rule problem, with 2 constants!

A trig. function of a constant is just a constant, so its derivative would be zero.

How deeply you must go is determined by how many "compositions" you have.

The two terms you began with have 3 compositions each. Let's look at the first term:

$a\cos^2(\pi w)=a(\cos(\pi w))^2$

If we let:

$f_1(x)=x^2$

$f_2(x)=\cos(x)$

$f_3(x)=\pi x$ then we may state:

$a(\cos(\pi w))^2=a\cdot f_1(f_2(f_3(w)))$

The chain rule tells us then:

$\frac{d}{dw}(a\cdot f_1(f_2(f_3(w))))=a\cdot f_1'(f_2(f_3(w)))\cdot f_2'(f_3(w))\cdot f_3'(w)$