Question

If `h(x)=sqrt(4+3f(x))` where f(1)=7 and f'(1)=4, how do you find h'(1)?

Answer 1

`h'(1)=6/5`

Explanation:

The given function is differentiated by using chain rule because it is a composite function of absolue value and polynomial function

Let
`u(x)=4+3f(x) and v(x)=sqrtx`

Then `color(blue)(h(x)=v(u(x)))`

`color(red)(h'(x)=v(u(x))')`

`color(red)(h'(x))=color(brown)(v'(u(x)))*color(green)(u'(x))`

Let us compute `color(red)(u'(x))` and `color(red)(v'(u(x)))`

`u(x)=4+3f(x)`
`color(green)(u'(x))=0+3f'(x)`
`color(green)(u'(x)=3f'(x))`

`v(x)=sqrtx`
`v'(x)=1/(2sqrtx)`

`color(brown)(v'(u(x)))=1/(2sqrt(u(x)))`

`color(brown)(v'(u(x))=1/(2sqrt(4+3f(x))))`

`color(red)(h'(x))=color(brown)(v'(u(x)))*color(green)(u'(x))`
`color(red)(h'(x))=color(brown)(1/(2sqrt(4+3f(x))))*color(green)(3f'(x))`

`h'(x)=(3f'(x))/(2sqrt(4+3f(x)))`

Then,
`h'(1)=(3f'(1))/(2sqrt(4+3f(1)))`

`h'(1)=(3*4)/(2sqrt(4+3*7))`
`h'(1)=12/(2sqrt25)=12/(2*5)=12/10=6/5`

Therefore `h'(1)=6/5`