The derivative of \(f(x)\) with respect to \(x\) is denoted by \(f'(x)\) where
$$
f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}.
$$
Equivalent notation for \(f'(x)\) include:
$$y’, \frac{df}{dx}, \frac{dy}{dx}, \frac{d}{dx}[f(x)]$$
$$
f'(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h}.
$$
Equivalent notation for \(f'(x)\) include:
$$y’, \frac{df}{dx}, \frac{dy}{dx}, \frac{d}{dx}[f(x)]$$
Basic Differentiation Rules: Let \(c\) be a constant.
- \( \displaystyle \frac{d}{dx}[c] = 0\)
- \( \displaystyle \frac{d}{dx}[x^n] = nx^{n-1}\)
- \( \displaystyle \frac{d}{dx}[e^x] = e^x\)
- \( \displaystyle \frac{d}{dx}[a^x] = a^x \ln a\)
- \( \displaystyle \frac{d}{dx}[\sin x] = \cos x\)
- \( \displaystyle \frac{d}{dx}[\csc x] = -\csc x \cot x\)
- \( \displaystyle \frac{d}{dx}[\cos x] = -\sin x\)
- \( \displaystyle \frac{d}{dx}[\sec x] = \sec x \tan x\)
- \( \displaystyle \frac{d}{dx}[\tan x] = \sec^2 x\)
- \( \displaystyle \frac{d}{dx}[\cot x] = -\csc^2 x\)
Derivative of a General Inverse Function:
$$
\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}
$$
$$
\frac{d}{dx}[f^{-1}(x)] = \frac{1}{f'(f^{-1}(x))}
$$
More Basic Differentiation Rules:
- \( \displaystyle \frac{d}{dx}[\ln x] = \frac{1}{x}\)
- \( \displaystyle \frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}\)
- \( \displaystyle \frac{d}{dx}[\sin^{-1} x] = \frac{1}{\sqrt{1-x^2}}\)
- \( \displaystyle \frac{d}{dx}[\sec^{-1} x] = \frac{1}{|x|\sqrt{x^2-1}}\)
- \( \displaystyle \frac{d}{dx}[\cos^{-1} x] = \frac{-1}{\sqrt{1-x^2}}\)
- \( \displaystyle \frac{d}{dx}[\csc^{-1} x] = \frac{-1}{|x|\sqrt{x^2-1}}\)
- \( \displaystyle \frac{d}{dx}[\tan^{-1} x] = \frac{1}{1+x^2}\)
- \( \displaystyle \frac{d}{dx}[\cot^{-1} x] = \frac{-1}{1+x^2}\)
Properties of Derivatives:If \(f(x)\) and \(g(x)\) are differentiable at \(x\) and \(c\) is a constant, then
- Sum/Difference Rule:
$$\frac{d}{dx}[f(x) \pm g(x)] = \frac{d}{dx}[f(x)] \pm \frac{d}{dx}[g(x)]$$ - Product Rule:
$$\frac{d}{dx}[f(x) \cdot g(x)] = \frac{d}{dx}[f(x)] \cdot g(x) + \frac{d}{dx}[g(x)] \cdot f(x) $$$$
\frac{d}{dx}[f(x) \cdot g(x)] = \frac{d}{dx}[f(x)] \cdot g(x) + \frac{d}{dx}[g(x)] \cdot f(x)
$$$$
\begin{aligned}
&\frac{d}{dx}[f(x) \cdot g(x)] \\
& \qquad = \frac{d}{dx}[f(x)] \cdot g(x) + \frac{d}{dx}[g(x)] \cdot f(x)
\end{aligned}
$$ - Quotient Rule:
$$\frac{d}{dx}\left[ \frac{f(x)}{g(x)}\right] = \frac{g(x) \cdot\frac{d}{dx}[f(x)] – \frac{d}{dx}[g(x)] \cdot f(x)}{(g(x))^2}$$$$\frac{d}{dx}\left[ \frac{f(x)}{g(x)}\right] = \frac{g(x) \cdot\frac{d}{dx}[f(x)] – \frac{d}{dx}[g(x)] \cdot f(x)}{(g(x))^2}$$$$
\begin{aligned}
& \frac{d}{dx}\left[ \frac{f(x)}{g(x)}\right] \\
& \qquad = \frac{g(x) \cdot\frac{d}{dx}[f(x)] – \frac{d}{dx}[g(x)] \cdot f(x)}{(g(x))^2}
\end{aligned}
$$ - Constant Multiplier Rule:
$$\frac{d}{dx}[c \cdot f(x)] = c \cdot \frac{d}{dx}[f(x)]$$ - Chain Rule: If, in addition, \(f(x)\) is differentiable at \(g(x)\) then
$$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot \frac{d}{dx}[g(x)]$$