Question

the derivative of ln x - e^-3x and ln x - (x^2 - 4)

Answer 1

Answer on Question#39986 – Math - Calculus

The derivative of $\ln x - e^{\wedge} - 3x$ and $\ln x - (x^{\wedge}2 - 4)$

Solution

For any functions $f$ and $g$ and any real numbers $a$ and $b$ the derivative of the function $h(x) = af(x) + bg(x)$ with respect to $x$ is:

$$h ^ {\prime} (x) = a f ^ {\prime} (x) + b g ^ {\prime} (x).$$

The sum rule

$$(f + g) ^ {\prime} = f ^ {\prime} + g ^ {\prime}$$

The subtraction rule

$$(f - g) ^ {\prime} = f ^ {\prime} - g ^ {\prime}.$$

If $f(x) = x^n$ , for any integer $n$ then

$$f ^ {\prime} (x) = n x ^ {n - 1}.$$ $$\frac {d}{d x} \left(e ^ {x}\right) = e ^ {x}$$ $$\frac {d}{d x} \left(\log_ {c} x\right) = \frac {1}{x \ln c}, \qquad c > 0, c \neq 1$$ $$(c) ^ {\prime} = 0$$

So,

1) $(\ln x - e^{-3x})' = (\ln x)' - (e^{-3x})' = \frac{1}{x*\ln e} - (-3e^{-3x}) = \frac{1}{x} + 3e^{-3x}$

2) $(\ln x - x^2 - 4)' = (\ln x)' - (x^2)' - 4' = \frac{1}{x} - 2*x^{2-1} - 0 = \frac{1}{x} - 2x^1 = \frac{1}{x} - 2x$