Find h'(3) if h=f o g, f(x)=7x^2+10 and the equation of the tangent line of g at 3 is y=−3x+4.

Question

anonymous · Answer

$h'(x)=f'(g(x))g'(x)$ is a start

anonymous · Answer

you are told that the equation of the tangent line of $g$ at $x=3$ is $y=-3x+4$ which is a round about way of telling you that $g'(3)=-3$

anonymous · Answer

it also tells you that $g(3)=-3\times 3+4=-5$

queelius · Answer

Tangent line: y = g(3) + g'(3)(x - 3); g'(3) = -3, and we are told y = -3x + 4, so g(3) + g'(3)(-3) = 4 (4 from the tangent line eq)... so, g(3) + 9 = 4, and therefore g(3) = 4 - 9 = -5. So, f'(g(x))g'(x) = 14(-5)(-3).

queelius · Answer

Looks like satellite was beating me to the punch, actually, but I had did my work on a sheet of paper before seeing his. :p

anonymous · Answer

now 
$$f'(x)=14x$$ and so $$f'(g(x))=14\times -5=-70$$ so now we are good to go 
$$h'(3)=f'(g(3))g'(3)=f'(-5)\times (-3)=-70\times (-3)$$ nope i got beat!

queelius · Answer

satellite showed his work much more clearly than I... well done.