1. ## Help with Derivative

I am having a few problems with two derivative problems.

The first problem is to find the derivative of sqr(1-25x^2)arcos(5x) and simplify if possible. I havent got the right answer after many tries i assumed you needed to use the product rule to compute this derivative.

The next problem is to find the tangent line at the point (-3sqr(3),1) for the curve x^(2/3) + y^(2/3) = 4.

If anyone could show me how to do this it would be much appreciated.

2. Originally Posted by NickytheNine
I am having a few problems with two derivative problems.

The first problem is to find the derivative of sqr(1-25x^2)arcos(5x) and simplify if possible. I havent got the right answer after many tries i assumed you needed to use the product rule to compute this derivative.
You will need product rule for this....

Note that $\frac{\,d}{\,dx}\left[\cos^{-1}u\right]=-\frac{\,d}{\,dx}\left[\sin^{-1}u\right]=-\frac{1}{\sqrt{1-u^2}}\cdot\frac{\,du}{\,dx}$

So when you apply product rule, you should end up with $-\frac{25}{\sqrt{1-25x^2}}\cos^{-1}(5x)+\sqrt{1-25x^2}\cdot\left[-\frac{5}{\sqrt{1-25x^2}}\right]$

Do you see why this is the case? I leave it for you to simplify this.

The next problem is to find the tangent line at the point (-3sqr(3),1) for the curve x^(2/3) + y^(2/3) = 4.

If anyone could show me how to do this it would be much appreciated.
For this one, use implicit differentiation...

You should end up with $\frac{\,dy}{\,dx}=-\frac{y^{\frac{1}{3}}}{x^{\frac{1}{3}}}$ [I leave it for you to show that this is the case].

Then plug your point in to find the slope of the tangent line. Then plug in all the known values into the point-slope equation for a line and get it into $y=mx+b$ form.

Can you try this one and see what you get?

--Chris