Thread: Find slope of the tangent line

Hi, I have no idea where to go for help and am hoping someone here can help me here for this problem.

The problem is based on sections 5.5(Bases Other than e and Applications)-5.6(Inverse Trigonometric Functions) in Calculus 9th edition by Larson. Chapter 5 is about Logarithmic, Exponential, and Other Transcendental Functions.

1. Find the slope of the tangent line to the graph of the given function at the given point.

y = 3x arcsin x , (1/2 , pi/4) Give the slope as an EXACT value, not a decimal approximation.

I have no idea where to start. Thank you.

Here is what I've done up to this point.

1. y= (3x)(arcsin(x)) ((1/2),(pi/4))

Found the derivative using the product rule: (f*g)' = f'*g + f*g'

Used d/dx (arcsin u) = (u')/(sqrt(1-u^2))

2. y' = (3)(arcsin(x)) + (3x)((1)/(sqrt(1-x^2)))

Simplified

3. y' = 3arcsin(x) + (3x)/(sqrt(1-x^2))

substituted x with 1/2 to get:

4. y = pi/2 + Sqrt(3)

Is pi/2 + sqrt(3) the correct answer?

Everything looks good to me!