Question

find the derivative of f(x) = ln x - e^3x and f(x) = ln x - ( x^2 - 4 )?

Answer 1

Answer on Question# 39985 - Math - Differential Calculus | Equations

Find the derivative of $f(x) = \ln x - e^{3x}$ and $f(x) = \ln x - (x^2 - 4)$.

Solution:

$$f(x) = \ln x - e^{3x}$$

Using: The sum rule:

$$(g \pm h)' = g' \pm h',$$

The chain rule:

If $(x) = f(h(x))$, then $\frac{dg}{dx} = \frac{dg}{dh} \cdot \frac{dh}{dx}$

Constant division rule: $\left(ag(x)\right)' = ag'(x)$

Power rule: $(x^n)' = n x^{n-1}$

As we know: $(\ln x)' = \frac{1}{x}$, $(e^x)' = e^x$.

Therefore, we obtain

$$f'(x) = (\ln x)' - e^{3x} \cdot (3x)' = \frac{1}{x} - 3e^{3x}.$$

Answer:

$$f'(x) = \frac{1}{x} - 3e^{3x}$$ $$f(x) = \ln x - (x^2 - 4)$$

Using: The sum rule: $(g \pm h)' = g' \pm h'$

$$(\ln x)' = \frac{1}{x}$$

Power rule: $(x^n)' = n x^{n-1}$

Constant rule: $(Const)' = 0$

Therefore, we obtain

$$f'(x) = (\ln x)' - (x^2)' + (4)' = \frac{1}{x} - 2x$$

Answer: $f'(x) = \frac{1}{x} - 2x$