Differentiating $f(x) = \ln5 + e^{25x} - e^3 +\dfrac{(7)}{\sqrt[3]{x}} - 6$

Question

Derivative of this logarithmic function. $$f(x) = \ln5 + e^{25x} - e^3 +\dfrac{(7)}{\sqrt[3]{x}} - 6$$

I got $$f'(x) = \dfrac{1}{5}+ e^{25x} (25) - e^3 -\dfrac{7}{3}\sqrt[3]{x^4}$$

is this correct?

Answer 1

The terms not containing $x$ is treated as constant. As you are finding derivative with respect to $x$.

$ln5$, $e^3$ and $6$ are constant terms. So derivative of these terms are zero.

Derivative of $\dfrac{(7)}{\sqrt[3]{x}}$

$$f(x)= 7 x^{\frac {-1}3}$$

$$f'(x) = 7 \cdot \frac {-1}3 x^{\frac{-4}3}$$

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

Answer 2

It's a bit wrong.

It should be $$f'(x) = e^{25x} \cdot(25) -\dfrac{7}{3\sqrt[3]{x^4}}$$ $e^3$ and $\ln5$ should be removed as they are constants.

and $\frac{d(\frac{1}{\sqrt[3]{x}})}{dx}=\frac{-1}{3\sqrt[3]{x^4}}$