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Qube
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Derivative of exponential function
f(x) = 4^x + e^(2tanx)
find f'(0)
I used log differentiation.
y = 4^x + e^(2tanx)
lny = xln4 + 2tanx(lne) = xln4 + 2tanx
y'/y = ln4 + 2sec^2(x)
Solving for y I get y = 2.
Plugging in x = 0 I get:
y'/2 = ln4 +2sec^(0) = ln4 + 2(1)
y' = 2(ln4+2)
This however, isn't the answer.
Homework Statement
f(x) = 4^x + e^(2tanx)
find f'(0)
Homework Equations
I used log differentiation.
y = 4^x + e^(2tanx)
lny = xln4 + 2tanx(lne) = xln4 + 2tanx
The Attempt at a Solution
y'/y = ln4 + 2sec^2(x)
Solving for y I get y = 2.
Plugging in x = 0 I get:
y'/2 = ln4 +2sec^(0) = ln4 + 2(1)
y' = 2(ln4+2)
This however, isn't the answer.
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