# Thread: Derivative of this logarithmic function.

1. ## Derivative of this logarithmic function.

f(x) = ln5 + e25x - e3 +(7)/(cubed root of x) - 6

f'(x) = (1)/(5) + e25x (25) - e3 - (7)/(3)(cubed root of x^4)

is this correct?

2. ## Re: Derivative of this logarithmic function.

Hey XxXScorpionXxX.

Since ln(5) is a constant the derivative of that term will be zero [as is another one].

Try using the power rule noting that 1/cube_root(x) = x^(-1/3).

3. ## Re: Derivative of this logarithmic function.

(Beaten to it).

4. ## Re: Derivative of this logarithmic function.

Originally Posted by XxXScorpionXxX
f(x) = ln5 + e25x - e3 +(7)/(cubed root of x) - 6

f'(x) = (1)/(5) + e25x (25) - e3 - (7)/(3)(cubed root of x^4)

is this correct?
no.

$\ln(5)$ is just a constant so $\dfrac{d}{dx} \ln(5) = 0$

same with $e^3$

$f(x) = \ln(5) + e^{25x} - e^3 + \dfrac {7}{x^{1/3}} - 6$

$f^\prime(x) = 25 e^{25x} -\dfrac 7 3 \dfrac{1}{x^{4/3}}$

5. ## Re: Derivative of this logarithmic function.

I see so 1/5 would be the derivative of ln5x then?

And I did try to use the power rule for 7 divided by the cubed root of x. does that not comout to -7 divided by 3 times cubed root of x^4?

6. ## Re: Derivative of this logarithmic function.

thanks for the clarification romsek you're always a great help.

7. ## Re: Derivative of this logarithmic function.

Originally Posted by XxXScorpionXxX
I see so 1/5 would be the derivative of ln5x then?
NO!

The derivative $\displaystyle D_x(\log(5x))=\dfrac{5}{5x}=\dfrac{1}{x}$.

8. ## Re: Derivative of this logarithmic function.

It's a bit wrong.

It should be
$$f'(x) = e^{25x} (25) -\dfrac{7}{3x\sqrt[3]{x}}$$
$e^3$ and $\ln5$ should be removed as they are constants.

and $f(x)=\dfrac{(7)}{\sqrt[3]{x}}\\f'(x)=\dfrac{-7}{3x^{4/3}}$