Derivatives with trig functions

Question

I am having trouble with a question on my homework and I have done some work but I'm not sure if I have completed all that is asked or how to take it further if needed.

Problem:

Let $\theta = \arccos(5x^3)$. Find $\frac{d\theta}{dx}$ in two ways:

(a) Use implicit differentiation starting from $\cos(\theta)= 5x^3$. Write your final answer in terms of $x$ by using SOH-CAH-TOA, a right triangle, and the Pythagorean Theorem.

(b) Use the Chain Rule extension of the formula $\frac{d}{dx} \arccos x = \frac{-1}{\sqrt{1-x^2}}$. Compare with your answer from part (a).

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Here is my work for part (a)

I drew a right triangle with an angle theta and labeled the hypotenuse as $1$ and the adjacent leg $5x^3$. Using the Pythagorean Theorem I found the opposite leg to be $\sqrt{1-25x^6}$. I'm not sure if this is all the question is asking or if I need to do something further. The wording of the problem is confusing me a little.

I also simply used this extra side to find $\sin(\theta) = \sqrt{1-25x^6}$.

Here is my work for part (b)

\begin{equation} \begin{split} \frac{d}{dx} \arccos 5x^3 &= \frac{-1}{\sqrt{1-(5x^3 )^2}} \frac{ d}{dx} (5x^3 )\\ &= \frac{-1}{\sqrt{1-25x^6}} 15x^2\\ &= \frac{-15x^2}{\sqrt{1-25x^6}}. \end{split} \end{equation}

If this is as far as I am supposed to take this problem are my answers supposed to be the same for part (a) and part (b)? If not then why not? I know I can simplify the answer for part (b) by rationalizing the denominator but it still wouldn't equal anything from part (a).

Any help would be greatly appreciated to help me finish this problem. I've looked at it several times for hours and am not sure where to go. Thanks!

Answer 1

For part a) you did the set up ok. Now if $cos\theta=5x^3$ then it follows $-{sin\theta}d\theta=15x^2dx$ and so $\frac{d\theta}{dx}=\frac{15x^2}{-sin\theta}$. Can you finish it from here?

Answer 2

For (a)...

Let $\cos \theta = 5x^3$; then by implicit differentiation we have $-\sin \theta \ \ d\theta/dx = 15x^2$. Dividing by $-\sin \theta$ we have $d\theta/dx = \frac {-15x^2}{\sin \theta}$; drawing the triangle with $1$ as the hypotenuse and $5x^3$ as the adjacent side, we have by the Pythagorean theorem $\sqrt{1-25x^6}$ as the opposite side, which is $\sin \theta$. Hence the complete answer is $$d\theta/dx = \frac {-15x^2}{\sqrt {1-25x^6}}$$

For (b)...

The derivative of $\arccos (x)$ is $d/dx \arccos (x) = \frac {-1}{\sqrt {{1-x^2}}}$; replacing $x$ with $5x^3$ and using the chain rule gives us $d \theta /dx \arccos (5x^3) = \frac {-1}{\sqrt {{1-(5x^3)^2}}}*15x^2$; simplifying gives us the same answer as (a).