Value of an implicit derivative

[Zack K]

Homework Statement

Find the value of h^'(0) if: $$h(x)+xcos(h(x))=x^2+3x+2/π$$

Homework Equations

Chain Rule
Product Rule

The Attempt at a Solution

I differentiated both sides, giving h'(x)+cos(h(x))-xh'(x)sin(h(x))=2x+3
Next I factored out and isolated h'(x) giving me: h'(x)=2x+3-cos(h(x))/1-sin(h(x))
My issue is that I do not know what h(0) is so I can't plug that into my equation to find h'(0)

 

[verty]

You are given that $h(x) + x\ \cos(h(x)) = x^2 + 3x + {2 \over \pi}$. This is true whatever we make x. What happens when x = 0?

 

[PeroK]

My issue is that I do not know what h(0) is so I can't plug that into my equation to find h^'0

Hmm. Normally, if you want to evaluate $h(0)$, you plug $0$ into an equation for $h(x)$. Compared to implicit differentiation, that would appear elementary!

 

[Zack K]

You are given that $h(x) + x\ \cos(h(x)) = x^2 + 3x + {2 \over \pi}$. This is true whatever we make x. What happens when x = 0?

Then h(x) is equal to 2/π?

 

[Mark44]

Then h(x) is equal to 2/π?

No, $h(0) = \frac 2 \pi$. You don't have an explicit formula for h(x).

Replace x with 0 in the formula you wrote in post #1, and maybe you can solve algebraically for h'(0).

I differentiated both sides, giving h'(x)+cos(h(x))-xh'(x)sin(h(x))=2x+3

Also, don't use the BBCode superscript icon for the prime symbol - ' Doing this makes it almost impossible to see.