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

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Answer 1 · 2016-10-25T18:46:07

#h'(1)=6/5#

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#