Basic Differentiation Rules [Aturan Dasar Turunan]

The derivative of a function $f$ is another function $f'$ [read "$f$ prime"] whose value at any number $x$ is $f'(x)=\lim\limits_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$.

If this limit does exist, we say that $f$ is differentiable at $x$. Finding a derivative is called differentiation; the part of calculus associated with the derivative is called differential calculus.

With Leibniz notation, introduced in the last section, we now have three notations for the derivative. If $y=f(x)$, we can denote the derivative of $f$ by $f'(x)$ or $D_{x}f(x)$ or $\dfrac{dy}{dx}$

Basic Differentiation Rules

$\dfrac{d}{dx}[c]=0$
$\dfrac{d}{dx}[x]=1$
$\dfrac{d}{dx}[cx^{n}]=cnx^{n-1}$
$\dfrac{d}{dx}[cu]=cu'$
$\dfrac{d}{dx}[u \pm v]=u' \pm v'$
$\dfrac{d}{dx}[u v]=u'v+uv'$
$\dfrac{d}{dx}[\dfrac{u}{v}]=\dfrac{u'v-uv'}{v^{2}}$
$\dfrac{d}{dx}[u^{n}]=nu^{n-1}u'$
$\dfrac{d}{dx}[\left |u \right |]=\dfrac{u}{\left |u \right |}(u'),\ u\neq 0 $
$\dfrac{d}{dx}[ln\ u]=\dfrac{u'}{u}$
$\dfrac{d}{dx}[e^{u}]=u'e^{u}$
$\dfrac{d}{dx}[log_{a}u]=\dfrac{u'}{(ln\ a)u}$
$\dfrac{d}{dx}[a^{u}]=a^{u}u'(ln\ a)$
$\dfrac{d}{dx}[sin\ u]=u'(cos\ u)$
$\dfrac{d}{dx}[cos\ u]=-u'(sin\ u)$
$\dfrac{d}{dx}[tan\ u]=u'(sec^{2}\ u)$
$\dfrac{d}{dx}[cot\ u]=-u'(csc^2\ u)$
$\dfrac{d}{dx}[sec\ u]=u'(sec\ u\ tan\ u)$
$\dfrac{d}{dx}[csc\ u]=-u'(csc\ u\ cot\ u)$
$\dfrac{d}{dx}[arcsin\ u]=\dfrac{u'}{\sqrt{1-u^{2}}}$
$\dfrac{d}{dx}[arccos\ u]=\dfrac{-u'}{\sqrt{1-u^{2}}}$
$\dfrac{d}{dx}[arctan\ u]=\dfrac{u'}{1+u^{2}}$
$\dfrac{d}{dx}[arccot\ u]=\dfrac{-u'}{1+u^{2}}$
$\dfrac{d}{dx}[arcsec\ u]=\dfrac{u'}{|u| \sqrt{u^{2}-1}}$
$\dfrac{d}{dx}[arccsc\ u]=\dfrac{-u'}{|u| \sqrt{u^{2}-1}}$
$\dfrac{d}{dx}[sinh\ u]=u'(cosh\ u)$
$\dfrac{d}{dx}[cosh\ u]=-u'(sinh\ u)$
$\dfrac{d}{dx}[tanh\ u]=u'(sech^{2}\ u)$
$\dfrac{d}{dx}[coth\ u]=-u'(csch^2\ u)$
$\dfrac{d}{dx}[sech\ u]=-u'(sech\ u\ tanh\ u)$
$\dfrac{d}{dx}[csch\ u]=-u'(csch\ u\ coth\ u)$
$\dfrac{d}{dx}[sinh^{-1}\ u]=\dfrac{u'}{\sqrt{u^{2}+1}}$
$\dfrac{d}{dx}[cosh^{-1}\ u]=\dfrac{u'}{\sqrt{u^{2}-1}}$
$\dfrac{d}{dx}[tanh^{-1}\ u]=\dfrac{u'}{1-u^{2}}$
$\dfrac{d}{dx}[coth^{-1}\ u]=\dfrac{u'}{1-u^{2}}$
$\dfrac{d}{dx}[sech^{-1}\ u]=\dfrac{-u'}{u\sqrt{1-u^{2}}}$
$\dfrac{d}{dx}[csch^{-1}\ u]=\dfrac{-u'}{|u| \sqrt{1+u^{2}}}$

Daftar bacaan:

Calculus Ninth Edition by Varberg, Purcel and Rigdon
Calculus Ninth Edition by Larson, and Edwards

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Basic Differentiation Rules [Aturan Dasar Turunan]

Basic Differentiation Rules

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