if h(x)=2g(x)-(1)/(g(x)), g(0)=2 and g'(0)=-1, find the exact value of h'(0). please help and show all steps!

Question

anonymous · Answer

For starters I would rewrite your h(x) function
$$h(x)=2g(x)-\frac{1}{g(x)}$$ using an exponent 

$$h(x)=2g(x)-(g(x))^{-1}$$.  Then, differentiate, using the chain rule on the second term.  

$$h'(x)=2g'(x)+(g(x))^{-2} \cdot g'(x)$$ 
so...

$$h'(x)=2g'(x)+\frac{g'(x)}{(g(x))^2}$$.

From here, you can just substitute zero for $$x$$ to get 

$$h'(0)=2g'(0)+\frac{g'(0)}{(g(0))^2}$$.  Substituting the values from the original problem will give you the final answer.