Proving C is a Differentiable Function: Inverse Function Theorem & Chain Rule

[kimsworld]

Homework Statement

Problem: Given C is the graph of the equation
2radical3 * sinpi(x)/3 =y^5+5y-3

Homework Equations

(1) Prove that as a set
C= {(x,y) Exists at all Real Numbers Squared | 2radical3 * sinpi(x)/3 =y^5+5y-3
is the graph of a function differentiable on all real numbers using the Inverse Function Theorem and Chain Rule.

(2) Obtain the slope and a cartesian equation of the line l tangent to C at the point (1,1). Prove your answer.

(3) A point (x(t), y(t)) moves all along C as t increases in all real numbers with constant horizontal rate with respect to t of 30 units per second
(For all t that exists as real numbers : x'(t) = 30/sec).
What is the vertical rate w.r.t. t when x(t) =1 ([y']x=1)?


*Note: The equation given for C cannot be solved for y using just the standard arithmetic operations, including radicals, but can be solved using inverse function notation.

The Attempt at a Solution

The extent of where I have gotten with the first proof is:

Let F(x)=G(y) ⇔ G^-1(F(x))=y
Because I do not know how to take the inverse of y^5+5y-3 I took the inverse of
2radical3 sinpi(x)/3

Thus,
F(y) = y^5+5y-3 and G(x)= 2radical3 sinpi(x)/3

G^-1 = 3/pi sin^-1 (x/2radical3)

With this I want to find that:

G^1(F(x)) = y= 3/pi sin^-1 [(x^5+5x-3)/ (2radical3)]

Now I want to prove that this is differentiable so I must take the derivative however this requires me to take the derivative of the inverse sin function and that is something learned in calculus 2. Does anybody know another way to go about this? Perhaps there is a simple way to take the inverse of y^5+5y-3 that I am overlooking?

 

[HallsofIvy]

"Implicit differentiation" is typically learned in Calculus I. Rather than solving for y, differentiate both sides of $2\sqrt{3} sin(\pi x)/3= y^5+ 5y- 3$ with respect to x.

 

[kimsworld]

"Implicit differentiation" is typically learned in Calculus I. Rather than solving for y, differentiate both sides of $2\sqrt{3} sin(\pi x)/3= y^5+ 5y- 3$ with respect to x.

So by using implicit differentiation I receive

y' = $\frac{2 \pi cos (\pi x/3)}{\sqrt{3}(5y^4+5)}$

Which is differentiable for all real numbers.

 

[SammyS]

So by using implicit differentiation I receive

$\displaystyle y' = \frac{2 \pi cos (\pi x/3)}{\sqrt{3}(5y^4+5)}$

Which is differentiable for all real numbers.

That looks good, provided that the original equation was

$\displaystyle 2\sqrt{3} \sin(\pi x/3)= y^5+ 5y- 3$​

The placement of the parenthesis is important.