Given
g
(
2
)
=
36
\displaystyle {g{{\left({2}\right)}}}={36}
g
(
2
)
=
36
,
g
′
(
2
)
=
1
\displaystyle {g}'{\left({2}\right)}={1}
g
′
(
2
)
=
1
and
f
(
x
)
=
g
(
x
)
\displaystyle {f{{\left({x}\right)}}}=\sqrt{{{g{{\left({x}\right)}}}}}
f
(
x
)
=
g
(
x
)
, find
f
′
(
2
)
\displaystyle {f}'{\left({2}\right)}
f
′
(
2
)
.
Preview
Question 6
Submit
Try a similar question
License
[more..]
\displaystyle
Enter your answer as a reduced fraction (like 5/3, not 10/6) or as an integer (like 4 or -2)
Do not enter mixed numbers
Enter DNE for Does Not Exist, oo for Infinity