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Derivatives Cheat Sheet

Function FormDerivative
Power Rule: y=cxn+Cy = c x^n + Cy=cnxn1y' = c n x^{n-1}
Sum/Difference Rule: y=u(x)±v(x)y = u(x) \pm v(x)y=u(x)±v(x)y' = u'(x) \pm v'(x)
Product Rule: y=u(x)v(x)y = u(x) \cdot v(x)y=u(x)v(x)+u(x)v(x)y' = u'(x) v(x) + u(x) v'(x)
Quotient Rule: y=u(x)v(x)y = \frac{u(x)}{v(x)}y=u(x)v(x)u(x)v(x)v2(x)y' = \frac{u'(x) v(x) - u(x) v'(x)}{v^2(x)}
General Power Rule: y=u(x)v(x)y = u(x)^{v(x)}y=uv(uvu+vlnu)y' = u^v \left( \frac{u' v}{u} + v' \ln u \right)

FunctionDerivative
y=exy = e^xy=exy' = e^x
y=ekxy = e^{kx}y=kekxy' = k e^{kx}
y=axy = a^xy=axlnay' = a^x \ln a
y=eu(x)y = e^{u(x)}y=u(x)eu(x)y' = u'(x) e^{u(x)} (Chain Rule)
FunctionDerivative
y=lnxy = \ln xy=1xy' = \frac{1}{x}
y=logaxy = \log_a xy=1xlnay' = \frac{1}{x \ln a}
y=lnu(x)y = \ln u(x)y=u(x)u(x)y' = \frac{u'(x)}{u(x)} (Logarithmic Differentiation)
FunctionDerivative
y=sinxy = \sin xy=cosxy' = \cos x
y=cosxy = \cos xy=sinxy' = -\sin x
y=tanxy = \tan xy=sec2x=1+tan2xy' = \sec^2 x = 1 + \tan^2 x
y=cotxy = \cot xy=csc2x=(1+cot2x)y' = -\csc^2 x = -(1 + \cot^2 x)
y=secxy = \sec xy=secxtanxy' = \sec x \tan x
y=cscxy = \csc xy=cscxcotxy' = -\csc x \cot x
FunctionDerivative
y=arcsinxy = \arcsin xy=11x2y' = \frac{1}{\sqrt{1 - x^2}}
y=arccosxy = \arccos xy=11x2y' = -\frac{1}{\sqrt{1 - x^2}}
y=arctanxy = \arctan xy=11+x2y' = \frac{1}{1 + x^2}
y=arccotxy = \operatorname{arccot} xy=11+x2y' = -\frac{1}{1 + x^2}
FunctionDerivative
y=sinhxy = \sinh xy=coshxy' = \cosh x
y=coshxy = \cosh xy=sinhxy' = \sinh x
y=tanhxy = \tanh xy=sech2xy' = \text{sech}^2 x
y=cothxy = \coth xy=csch2xy' = -\text{csch}^2 x

Chain Rule (Derivative of a Composite Function)

Section titled “Chain Rule (Derivative of a Composite Function)”

If y=f(u(x))y = f(u(x)), then:

dydx=dydududx=f(u)u(x)\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = f'(u) \cdot u'(x)

Given x=f(t),y=g(t)x = f(t), y = g(t):

dydx=y˙x˙=g(t)f(t)\frac{dy}{dx} = \frac{\dot{y}}{\dot{x}} = \frac{g'(t)}{f'(t)}

Second derivative:

d2ydx2=x˙y¨y˙x¨x˙3\frac{d^2y}{dx^2} = \frac{\dot{x} \ddot{y} - \dot{y} \ddot{x}}{\dot{x}^3}

For equations where yy is not explicitly expressed in terms of xx, differentiate both sides and solve for yy'.

Example:
For x2+y2=r2x^2 + y^2 = r^2,

2x+2ydydx=0    dydx=xy2x + 2y \frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y}

If y=f(x)y = f(x) has an inverse x=f1(y)x = f^{-1}(y), then:

dydx=1dxdy\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}}

Example:
For y=arccosxy = \arccos x,

x=cosy    dxdy=siny    dydx=11x2x = \cos y \implies \frac{dx}{dy} = -\sin y \implies \frac{dy}{dx} = -\frac{1}{\sqrt{1 - x^2}}