Calculus/Differentiation/Solutions

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< Calculus | Differentiation(Redirected from Calculus/Answers to introductory concepts)

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[edit] Find The Derivative By Definition

1. 2x

$f(x)\,$	= $x^2\,$
	= $\lim_{\Delta x \to 0}\frac{(x+\Delta x)^2-x^2}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{x^2+2x\Delta x+\Delta x^2-x^2}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{2x\Delta x+\Delta x^2}{\Delta x}$
	= $\lim_{\Delta x \to 0}2x+\Delta x$
	= $2x\,$

2. 2

$f(x)\,$	= $2x+2\,$
$f'(x)\,$	= $\lim_{\Delta x \to 0}\frac{[2(x+\Delta x) + 2] - (2x + 2)}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{2x+2\Delta x + 2 - 2x - 2}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{2\Delta x}{\Delta x}$
	= $2\,$

3. x

$f(x)\,$	= $\frac{1}{2}x^2\,$
$f'(x)\,$	= $\lim_{\Delta x \to 0}\frac{\frac{1}{2}(x+\Delta x)^2-\frac{1}{2}x^2}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{\frac{1}{2}(x^2+2x\Delta x+\Delta x^2)-\frac{1}{2}x^2}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{\frac{x^2}{2}+\frac{2x \Delta x}{2}+\frac{\Delta x^2}{2}-\frac{x^2}{2}}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{2x\Delta x + \Delta x^2}{2\Delta x}$
	= $\lim_{\Delta x \to 0}x+\Delta x$
	= $x\,$

4. 4x + 4

$f(x)\,$	= $2x^2+4x+4\,$
$f'(x)\,$	= $\lim_{\Delta x \to 0}\frac{[2(x+\Delta x)^2 + 4(x+\Delta x)+4] - (2x^2+4x+4)}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{2(x^2+2x\Delta x + \Delta x^2) + 4x + 4\Delta x + 4 - 2x^2 - 4x - 4}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{2x^2 + 4x\Delta x + 2\Delta x^2 + 4\Delta x - 2x^2}{\Delta x}$
	= $\lim_{\Delta x \to 0}\frac{4x\Delta x + 2\Delta x^2 + 4\Delta x}{\Delta x}$
	= $\lim_{\Delta x \to 0}4x+2\Delta x + 4$
	= $4x + 4\,$

[edit] Prove Differentiation Rules

[edit] Proof of the Derivative of a Constant Function

$\mbox{If } c = f(x) \mbox{, then}\,$
$\frac{d}{dx} \left[\,c\right]$	$= f'(x)\,$
	$= \lim_{\Delta x \to 0} { f(x+\Delta x)-f(x) \over \Delta x }$
	$= \lim_{\Delta x \to 0} { c-c \over \Delta x }$
	$= \lim_{\Delta x \to 0} \,0$
	$= 0\,$

[edit] Proof of the Derivative of a Linear Function

$\mbox{If } mx+b = f(x) \mbox{, then}\,$
$\frac{d}{dx} (mx+b)$	$= f'(x)\,$
	$= \lim_{\Delta x \to 0} { f(x+\Delta x)-f(x) \over \Delta x }$
	$= \lim_{\Delta x \to 0} { [m(x+\Delta x)+b]-[mx+b] \over \Delta x }$
	$= \lim_{\Delta x \to 0} { mx+m\Delta x+b-mx-b \over \Delta x }$
	$= \lim_{\Delta x \to 0} { m\Delta x \over \Delta x }$
	$= \lim_{\Delta x \to 0} \,m$
	$= m\,$

[edit] Proof of the Constant Multiple Rule

$\frac{d}{dx}\left[cf(x)\right]$	= $\lim_{\Delta x \to 0}\frac{cf\left(x+\Delta x \right)-cf\left(x\right)}{\Delta x}$
	= $c\lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$
	= $c\frac{d}{dx}\left[f(x)\right]\,$

[edit] Proof of the Addition and Subtraction Rules

$\frac{d}{dx}\left[f(x)\pm g(x)\right]$	= $\lim_{\Delta x\to0}\frac{\left[f(x+\Delta x)+g(x+\Delta x)\right]-\left[f(x)+g(x)\right]}{\Delta x}$
	= $\lim_{\Delta x\to 0}\frac{f(x+\Delta x)+g(x+\Delta x)-f(x)-g(x)}{\Delta x}$
	= $\lim_{\Delta x\to 0}\left[\frac{f(x+\Delta x)-f(x)}{\Delta x}+\frac{g(x+\Delta x)-g(x)}{\Delta x}\right]$
	= $\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}+\lim_{\Delta x\to 0}\frac{g(x+\Delta x)-g(x)}{\Delta x}$
	= $\frac{d}{dx}\left[f(x)\right]+\frac{d}{dx}\left[g(x)\right]$

[edit] Find The Derivative By Rules

$4x\,$
$\frac{1}{\sqrt[3]{x^2}}\,$
$\frac{1}{\sqrt[3]{x^2}}-\frac{2}{x^3}\,$
$\frac{1}{x}-2e^x+\frac{1}{2\sqrt{x}}\,$
$\cos (x) - \sin (x)\,$
$2(x+5)\,$
$\frac{x}{\sqrt{1+x^2}}\,$
$\frac{x^3}{\sqrt{y}}+6\sqrt{y}x^2\,$
$\frac{-8\sqrt{x}}{y^5}+\frac{1}{y^4\sqrt{x}}\,$
$2\sqrt{2x^2+1}(3y^4+2y)(12y^3+2)+\frac{2x(3y^4+2y)^2}{\sqrt{2x^2+1}}\,$
$\ln (4)4^x\,$
$\frac{2^{x-3}9x^2}{2\sqrt{x^3-2}}+3\ln (2)\sqrt{x^3-2}(2^{x-3})+\frac{1}{x}\,$
$4xe^xy^2z^3+2xe^xyz^4+(xe^xy^2z^4+e^xy^2z^4)\,$
$\frac{1}{x\ln 4}+\frac{2}{x}\,$
$3e^x+4\sin (x) - \frac{1}{4x}\,$
$10 x 4 + 16 x + 1$
$49 x 6 + 40 x 4 + 3 x 2 + 2 x - 1$

[edit] Implicit Differentiation

Recall that $\frac{dy}{dx}$ is the same as $y'$ .

1.

$\frac{d}{dx} (x^2 + y^2 = 1)$
$2x + 2y\frac{dy}{dx} = 0$
solve for $\frac{dy}{dx}$
$\frac{dy}{dx} = -\frac{x}{y}$

2.

$\frac{d}{dx} (x^3 + y^3 = xy)$
$3x^2 + 3y^2 \frac{dy}{dx} = y + x \frac{dy}{dx}$
solve for $\frac{dy}{dx}$
$3x^2 - y = \frac{dy}{dx} (x-3y^2)$
$\frac{dy}{dx} = \frac{3x^2 - y}{x-3y^2}$