Math 22 The Derivative and the Slope of a Graph

Slope of a Graph

We can estimate the slope at the given point to be

Slope = ${\frac {\Delta y}{\Delta x}}={\frac {\text{change in y}}{\text{change in x}}}$

Difference Quotient

 The slope $m$ of the graph of $f$ at the point $(x,f(x))$ can be 
 written as :
 
 $m=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$
 
 The right side of this equation ${\frac {f(x+h)-f(x)}{h}}$ is called Difference Quotient

Example: Find the Different Quotient of

1) $f(x)=x^{2}-1$

Solution: Consider ${\frac {f(x+h)-f(x)}{h}}={\frac {(x+h)^{2}-1-(x^{2}-1)}{h}}$

$={\frac {x^{2}+2xh+h^{2}-1-x^{2}+1}{h}}={\frac {2xh+h^{2}}{h}}={\frac {h(2x+h)}{h}}=2xh$

2) $f(x)=4x-1$

Solution:
Consider ${\frac {f(x+h)-f(x)}{h}}={\frac {4(x+h)-1-(4x-1)}{h}}={\frac {4x+4h-1+4x+1}{h}}={\frac {4h}{h}}=4$

Definition of the Derivattive

 The derivative of $f$ at $x$ is given by
 
 $f'(x)=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}$
 
 provided this limit exists. A function is differentiable at Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x}
 when its 
 derivative exists at . The process of finding derivatives is called differentiation.

Example: Use limit definition to find the derivative of

1) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=x^2+2x}

Solution: Consider: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h\to 0} \frac {(x+h)^2+2(x+h)-(x^2+2x)}{h}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\lim_{h\to 0} \frac {x^2+2xh+h^2 +2x+2h-x^2-2x}{h}=\lim_{h\to 0} \frac {2xh+h^2+2h}{h}}

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{h\to 0} \frac{h(2x+h+2)}{h}=\lim_{h\to 0} (2x+h+2)=2x+(0)+2=2x+2}

2) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)=2x^2-3x+1}

Solution:
Consider: Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=\lim_{h\to 0} \frac {2(x+h)^2-3(x+h)+1-(x^2-3x+1)}{h}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle =\lim_{h\to 0} \frac {2x^2+4xh+2h^2 -3x-3h+1-2x^2+3x-1}{h}=\lim_{h\to 0} \frac {4xh+2h^2-3h}{h}}
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{h\to 0} \frac{h(4x+2h-3)}{h}=\lim_{h\to 0} (4x+2h-3)=4x+2(0)-3=4x-2}

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