Find the derivative of y=f(g(sin(x))) evaluated at x=0

Question

use the following table to answer the question

Function | f(x) | f'(x) | g(x) | g'(x) |

x=0 | 2 | 5 | 0 | 2 |

x=1 | 1 | -2 | 3 | 0 |

x=2 | 4 | 4 | 1 | -1 |

I did my best to make the graph any help would be greatly appreciated and I believe the chain rule is what will be used here :)

Answer 1

Daniel, you are correct that chain rule will most definitely be used in the problem solution. Granted, it is a weird nested chain rule I can confidently say I haven't handled yet, or at least having a trigonometric function on the inner-term.

The general equation for nested chain rule for n-nested functions is as follows:

y = f₁ ( f₂ ( f₃ ( ... f_n ( x ) ... )

dy/dx = y' = f₁' ( f₂ ( f₃ ( ... f_n ( x ) ... ) * f₂' ( f₃ ( ... f_n ( x ) ... ) * ... f_n' ( x )

You can run through the individual steps of the logic of the above general equation but it will work for this problem.

y = f ( g ( sin ( x ) ) ) ⇒ dy/dx = y' = f ' ( g ( sin ( x ) ) ) * g' ( sin ( x ) ) * cos ( x ).

y' ( 0 ) = f' ( g ( sin ( 0 ) ) ) * g' ( sin ( 0 ) ) * cos ( 0 )

y' ( 0 ) = f' ( g ( 0 ) ) * g' ( 0 ) * 1

y' ( 0 ) = f' ( 0 ) * 2

y' ( 0 ) = 2 * 5

∴ y' ( 0 ) = 10

I hope this helps! Let me know if you have questions in the comments!