Implicit differentiation

[Rumor]

Here's the problem:

"Consider the curve defined by the implicit equation y+5cosy=x^2+2.

a) Find dy/dx in terms of x and y.

b) Find the equation of the tangent line at the point shown where x=2.

c) Find two positive values of x in the portion of the curve shown at which there is a vertical tangent line." (There's a picture here of the graph, given so that one could visually look to see where they think a vertical tangent line might be.) "Use calculus to find the values to 3 decimal places. You can use your calculator to solve the equation that must be true for a vertical tangent line."

Any help on how to solve these, please?

 

[Deleted member 4993]

Here's the problem:

"Consider the curve defined by the implicit equation y+5cosy=x^2+2.

a) Find dy/dx in terms of x and y.

b) Find the equation of the tangent line at the point shown where x=2.

c) Find two positive values of x in the portion of the curve shown at which there is a vertical tangent line." (There's a picture here of the graph, given so that one could visually look to see where they think a vertical tangent line might be.) "Use calculus to find the values to 3 decimal places. You can use your calculator to solve the equation that must be true for a vertical tangent line."

Any help on how to solve these, please?

This problem is very similar to that of:

viewtopic.php?f=3&t=37219

Where are you stuck?

 

[BigGlenntheHeavy]

\(\displaystyle y+5cos(y)=x^{2}+2\)

\(\displaystyle y'-5sin(y)y'=2x \ \implies \ y' \ = \ \frac{2x}{1-5sin(y)}, \ Given \ x=2, \ we \ need \ a \ y \ to \ find \ y'.\)

\(\displaystyle Ergo, \ \ y+5cos(y)=2^{2}+2=6 \ \implies \ y \ = \ 4.928 \ (Use \ trusty \ TI-89 \ or \ grunt \ it \ out.)\)

\(\displaystyle Hence, \ (2,4.928), \ m \ = \ .67978, \ therefore \ y-4.928 \ = \ .67978(x-2) \ \implies \ y \ = \ .67978x+3.56844.\)

\(\displaystyle The \ above \ is \ the \ line \ tangent \ at \ f(x,y) \ at \ 2 \ (black \ line)\)

\(\displaystyle Vertical \ tangent \ line: \ 1-5sin(y) \ = \ 0 \ \implies \ y \ = \ arcsin(1/5).\)

\(\displaystyle Therefore, \ x \ = \ [arcsin(1/5)+5cos[arcsin(1/5)]-2]^{1/2} \ (green \ line).\)

\(\displaystyle I'll \ leave \ it \ to \ you \ to \ find \ another \ vertical \ tangent \ line.\)

[attachment=0:3fkzrp0r]odd.jpg[/attachment:3fkzrp0r]