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