What Is the Correct Derivative of f(x) = 4^x + e^(2tanx) at x=0?

[Qube]

Derivative of exponential function

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.

 

[Mark44]

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.

No - don't do this yet. Solve for y' algebraically.

There's a difference between y' and y'(0). The first is a function, and the second is a number. Your need the function.

Plugging in x = 0 I get:

y'/2 = ln4 +2sec^(0) = ln4 + 2(1)

y' = 2(ln4+2)

This however, isn't the answer.

 

[Mark44]

Btw, your title is misleading. 4^x is NOT a power function. It's an exponential function.

In a power function, the variable is in the base, and the exponent is constant. E.g., x³, y⁵, and so on are power functions.

 

[Qube]

No - don't do this yet. Solve for y' algebraically.

There's a difference between y' and y'(0). The first is a function, and the second is a number. Your need the function.

Alright.

y' = y(ln4 + 2sec^2(x))

How do I proceed? Do I plug in y =2? If so I still get the same answer if I plug in x = 0.

 

[LCKurtz]

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

This step in incorrect. $\ln(a+b) \ne \ln a + \ln b$.

 

[Qube]

I got it now! Thanks! I should probably review my log rules. It's amazing that after two years of calculus and a 5 on both AP Calc AB and BC I still make the simplest of pre-calculus mistakes!

 

[Ray Vickson]

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.

Your second line essentially says that
\log(a+b) = \log(a) + \log(b)\;\leftarrow \text{ false!}
What IS true is $\log(a\cdot b) = \log(a) + \log(b)$.

 

[Mark44]

This step in incorrect. $\ln(a+b) \ne \ln a + \ln b$.

I glossed right over that in the OP's work. Oh, well.