Am I not seeing something here?

[MathNoob94]

Hi,
So I have a question that asks me to find first and second derivative of h of x I get theta(2cos(theta)) but the answer is not correct. Question 29.

I have attached pictures to help explain my issue.

Thank you!

 

[stapel]

Maybe you could provide images that aren't sideways...?

 

[MathNoob94]

Sorry lol I took it from my phone. But I managed to figure out the question. However I had a new question regarding chain rule it was f(x) = x^2 sqrt(3-2x) I'm getting confused on what to do exactly. I did the product rule then afterwards I get stumped.

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[ksdhart]

Your work for the first derivative looks good to me, but the second derivative is a little off. You are given: \(\displaystyle h\left(\theta \right)=\theta sin\left(\theta \right)\)

So, by the product rule, the first derivative is: \(\displaystyle \frac{dh}{d\theta }=\theta \left(cos\left(\theta \right)+sin\left(\theta \right)\right)\)

That's all good so far. Then when taking the second derivative, you establish: \(\displaystyle f\left(\theta \right)=\theta\) and \(\displaystyle g\left(\theta \right)=sin\left(\theta \right)+cos\left(\theta \right)\)

The problem is that the first term of the second derivative (fg') is going to be \(\displaystyle \theta \left(\frac{d}{d\theta}cos\left(\theta \right)+\frac{d}{d\theta }sin\left(\theta \right)\right)\)

Theta is supposed to be multiplied by the entirety of the derivative of g. You only multiplied by the first term, the cosine derivative.

 

[stapel]

I had a new question regarding chain rule it was f(x) = x^2 sqrt(3-2x)

I will guess that the above means the following:

I am given the function \(\displaystyle \,f(x)\, =\, x^2\, \sqrt{3\, -\, 2x\,}.\) The instructions say to find the derivative of this function by using the Chain Rule.

Would this interpretation be correct?

I'm getting confused on what to do exactly.

You are supposed to apply the Chain Rule to a variable-containing part of the function which lies within another operation. In this case, you have a polynomial inside a square root (or, which is the same thing, a one-half power). You have to differentiate the one-half power before you can get "inside" to differentiate the polynomial.

Has your class not covered the Chain Rule yet?

I did the product rule then afterwards I get stumped.

Unfortunately, since we cannot see what you did, we cannot advise what to do next, better, etc. Kindly please review this posting and then reply with the requested information. Thank you!