Let f and g be differentiable functions with the following values:
f(1) = 2 f'(1) = 3 g(1) = 2 g'(1) = -3
f(2) = 4 f'(2) = -4 g(2) = -5 g'(2) = 5
If h(x)=1/(g(f(x))), write the equation of the line tangent to h at x =1

so far, I've got
$$h'(x) = \frac{ -(g'(f(x))*f'(x) }{ (g(f(x)))^{2}}$$ and
$$h'(1) = -\frac{ 3 }{ 5 }$$ I am not sure how to get the equation of the line though...

Question

Let f and g be differentiable functions with the following values:
f(1) = 2     f'(1) = 3      g(1) = 2     g'(1) = -3
f(2) = 4     f'(2) = -4    g(2) = -5    g'(2) = 5
If h(x)=1/(g(f(x))), write the equation of the line tangent to h at x =1



so far, I've got
$$h'(x) = \frac{ -(g'(f(x))*f'(x) }{ (g(f(x)))^{2}}$$ and
$$h'(1) = -\frac{ 3 }{ 5 }$$ I am not sure how to get the equation of the line though...

hartnn · Answer

to find equation of a line, you'll need a slope and a point
slope = h'(1) = -3/5 
find the point,
when x= 1, h(1) =... ?

hartnn · Answer

then the point is just (1,h(1))

jigglypuff314 · Answer

oh, okay thank you!
I get it now :D

hartnn · Answer

welcome ^_^