Differentiat the following: f(x)= ln((e^(3x)+2)/(e^(5x))

Question

Having trouble with some calc 2 homework. Differentiating logs is proving to be difficult for me.

Answer 1

Isn’t it fun to see different ways to solve a problem! Interested in another? How about using the equivalent exponential statement to apply differentiation to and avoid differentiation of the logarithmic form all together. Sound fun?

Okay first recall that

log_B(N) = E If and only if B^E=N.

Next recall

ln(N) = log_e(N).

So

f(x) = log_e((e^3x+2)/e^5x) ⇔ e^f(x) = (e^3x+2)/e^5x

or

e^f(x) = e^-2x + 2e^-5x

⇒ d / dx (e^f(x)) = d / dx (e^-2x + 2e^-5x)

Next recall d / dx (e^u(x)) = e^u(x) * d / dx (u(x)).

Therefore,

d / dx (e^f(x)) = d / dx (e^-2x + 2e^-5x)

⇒ e^f(x) d / dx (f(x)) = -2e^-2x -10e^-5x.

⇒ d /dx (f(x)) = ( -2e^-2x -10e^-5x ) / e^f(x).

But we have an equivalent form of e^{f(x) =} e^-2x + 2e^-5x.

Hence d / dx (f(x)) = f’(x) = (-2e^-2x -10e^-5x) / (e^-2x + 2e^-5x).

The art of math is in the path.

Answer 2

desmos.com/calculator/ddddu17ub5

Answer 3

ln [ (2+ e^3x) / e^5x ] = ln (2e^-5x + e^-2x)

Using chain rule, differentiating ln p gives 1/p dp/dx so applying this here gives

(-10 e^-5x -2 e ^-2x) / (2e ^-5x + e^-2x)

Mike