If 2f'(1) = f(1), then the derivative of log[f(e*)] at x = 0 is =
Answers
Answered by
0
Step-by-step explanation:
Let y=logf(e
x
)
∴
dx
dy
=
f(e
x
)
1
×f
′
(e
x
).e
x
=
f(e
z
)
e
z
.f
′
(e
z
)
(
dx
dy
)
z−0
=
f(e
0
)
e
0
.f
′
(e
0
)
=
f(1)
1.f
′
(1)
=
4
2
=
2
1
Answered by
0
Answer:
1/2
Step-by-step explanation:
2f'(1)=f(1)
y = log f(e^x)
dy/dx = 1/f(e^x) X f'(e^x) X e^x
dy/ dx = e^x. f'(e^x) /f(e^x)
dy / dx = e^0 f'(e^0) /f(e^0)
dy/ dx = 1. f'(1)/2f'(1)
So, dy/dx = 1/2 (as f'(1) cancel to f'(1))
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