1. ## Chain Rule help

I'm reviewing and studying for finals and I've run into some trouble with the chain rule.

I have no problem using the chain rule to differentiate problems like this:

y = (3x + 1)^2
y'= 2(3x + 1)(3) = 6(3x +1)

or this

y = 2^cotx
lny = ln(2^cotx) = cotx*ln2
y'/y = cotx(1/2*0) + ln2*-csc^2x
y'= y(-ln2*csc^2x)
y'= -2^cotx(ln2*csc^2x)

But when I comes across problems like this, using the chain rule becomes gets confusing:

y = 2^sin3.14x

I don't know how to differentiate sin3.14x because I can't distinguish between the first and second term. Is the first term sin3.14 and the second x, is the first term sin and the second 3.14x, or should it all be treated as one term?

Anyway, this how I have tried to slove it:

lny = ln(2^sin3.14x) = sin3.14x*ln2
y'/y = sin3.14x(1/2*0) + ln2(sin3.14 + cos3.14x)
y' = y(ln2(sin3.14 + cos3.14x)
y' = 2^sin3.14x(ln2(sin3.14 + cos3.14x))

It's supposed to be y' = 2^sin3.14x(3.14ln2)cos3.14x

Will someone kindly tell me what I'm doing wrong?

Thanks

2. Originally Posted by zachb
I'm reviewing and studying for finals and I've run into some trouble with the chain rule.

I have no problem using the chain rule to differentiate problems like this:

y = (3x + 1)^2
y'= 2(3x + 1)(3) = 6(3x +1)

or this

y = 2^cotx
lny = ln(2^cotx) = cotx*ln2
y'/y = cotx(1/2*0) + ln2*-csc^2x
y'= y(-ln2*csc^2x)
y'= -2^cotx(ln2*csc^2x)

But when I comes across problems like this, using the chain rule becomes gets confusing:

y = 2^sin3.14x

I don't know how to differentiate sin3.14x because I can't distinguish between the first and second term. Is the first term sin3.14 and the second x, is the first term sin and the second 3.14x, or should it all be treated as one term?

Anyway, this how I have tried to slove it:

lny = ln(2^sin3.14x) = sin3.14x*ln2
y'/y = sin3.14x(1/2*0) + ln2(sin3.14 + cos3.14x)
y' = y(ln2(sin3.14 + cos3.14x)
y' = 2^sin3.14x(ln2(sin3.14 + cos3.14x))

It's supposed to be y' = 2^sin3.14x(3.14ln2)cos3.14x

Will someone kindly tell me what I'm doing wrong?

Thanks
This is why you need to use parenthesis:
"sin3.14x" should be sin(3.14x)

Both the 3.14 and the x are inside the sine function.

Also, when you differentiate this expression, ln(2) is simply a constant, so you don't need to take the derivative with respect to it.

Here's what I would do with it:
ln(y) = ln(2^{sin(3.14x)}) = (ln(2)) * sin(3.14x)
y'/y = (ln2) * (cos(3.14x) * 3.14)

y'/y = 3.14*ln(2)*cos(3.14x)

y' = y * [3.14*ln(2)*cos(3.14x)]

y' = 2^{sin(3.14x)} * 3.14*ln(2)*cos(3.14x)

-Dan

3. This is why you need to use parenthesis.
For some reason my book doesn't use parenthesis for functions like that.