Reference
Formula: \[f’(x)=\mathop {\lim}\limits_{h \to 0} \frac{f\left(x+h\right)-f\left(x\right)}{h}\] \[\] An alternate form of the derivative: \[f’(a)=\mathop {\lim}\limits_{x \to a}\frac{f\left(x\right)-f\left(a\right)}{x-a}\]\[\]Point-slope form line equation: \[y-y_{1}=m\left(x-x_{1}\right)\] Tangent Line: a line that touches the curve at a point where the curve's instantaneous rate of change is equivalent to the slope of the line.
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EX 1
Find the derivative of \(f\left(x\right)=x^{2}+4x+4\).
Using the formula:
\[f’(x)=\mathop {\lim}\limits_{h \to 0}\frac{\left(x+h\right)^{2}+4\left(x+h\right)+4-x^{2}-4x-4}{h}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{x^{2}+2xh+h^{2}+4x+4h+4-x^{2}-4x-4}{h}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{2xh+h^{2}+4h}{h}=\mathop {\lim}\limits_{h \to 0}(2x+h+4)=2x+4\]
EX 2
Find \(f’(x)\) given that \(f(x)=\frac{4}{x}\).
Using the formula:
\[f’(x)= \mathop {\lim}\limits_{h \to 0} \frac{\frac{4}{x+h}-\frac{4}{x}}{h}=\] \[\mathop {\lim}\limits_{h \to 0}\frac{\frac{4\left(x\right)}{\left(x+h\right)\left(x\right)}-\frac{4\left(x+h\right)}{\left(x+h\right)\left(x\right)}}{h}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{\frac{4\left(x\right)-4\left(x+h\right)}{\left(x+h\right)\left(x\right)}}{h}=\mathop {\lim}\limits_{h \to 0}\frac{4\left(x\right)-4\left(x+h\right)}{\left(x+h\right)\left(x\right)\left(h\right)}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{4x-4x+4h}{\left(x+h\right)\left(x\right)\left(h\right)}=\mathop {\lim}\limits_{h \to 0}\frac{4h}{\left(x+h\right)\left(x\right)\left(h\right)}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{4}{\left(x+h\right)\left(x\right)}=\mathop {\lim}\limits_{h \to 0}\frac{4}{x^{2}+h}=\frac{4}{x^{2}}\]
EX 3
Find \(f'\left(4\right)\) given that \(f(x)=\sqrt{x}\).
Using the formula:
\[f’(x)= \mathop {\lim}\limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\]\[\mathop {\lim}\limits_{h \to 0} \frac{\left(\sqrt{x+h}-\sqrt{x}\right)\left(\sqrt{x+h}+\sqrt{x}\right)}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\] \[\mathop {\lim}\limits_{h \to 0} \frac{x+h-x}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\mathop {\lim}\limits_{h \to 0} \frac{h}{h\left(\sqrt{x+h}+\sqrt{x}\right)}=\] \[\mathop {\lim}\limits_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}=\frac{1}{\sqrt{x}+\sqrt{x}}=\frac{1}{2\sqrt{x}}\]\[\therefore\] \[f’(4)=\frac{1}{2\sqrt{4}}=\frac{1}{2}\]
EX 4
Find the equation of the tangent line to the parabola \(y=2x^{2}-5x+3\) at the point \(x=1\).
- Find the derivative \[f'(x)=\mathop {\lim}\limits_{h \to 0}\frac{2\left(x+h\right)^{2}-5\left(x+h\right)+3-2x^{2}+5x-3}{h}=\] \[\mathop {\lim}\limits_{h \to 0}\frac{2x^{2}+4xh+2h^{2}-5x-5h+3-2x^{2}+5x-3}{h}=\] \[\mathop {\lim}\limits_{h \to 0}\frac{4xh+2h^{2}-5h}{h}=\mathop {\lim}\limits_{h \to 0}(4x+2h-5)=4x-5\]
- Substitute the \(x\)-coordinate into the derivative to find \(m\) \[m=f'(1)=4-5=-1\]
- Find the \(y\)-coordinate \[f(1)=2-5+3=0\]
- Write the equation of the tangent line at the point \[y-0=-1\left(x-1\right)\]\[y=-x+1\]
EX 5
Find the equation of the tangent line to \(y=\frac{5}{1+x^{2}}\) at the point \(x=2\).
- Find the derivative \[f'\left(x\right)=\mathop {\lim}\limits_{h \to 0}\frac{\frac{5}{1+\left(x+h\right)^{2}}-\frac{5}{1+x^{2}}}{h}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{\frac{5}{1+x^{2}+2xh+h^{2}}-\frac{5}{1+x^{2}}}{h}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{\frac{5\left(1+x^{2}\right)-5\left(1+x^{2}+2xh+h^{2}\right)}{\left(1+x^{2}+2xh+h^{2}\right)\left(1+x^{2}\right)}}{h}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{5+5x^{2}-5-5x^{2}-10xh-5h^{2}}{\left(1+x^{2}+2xh+h^{2}\right)\left(1+x^{2}\right)h}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{-10xh-5h^{2}}{\left(1+x^{2}+2xh+h^{2}\right)\left(1+x^{2}\right)h}=\]\[\mathop {\lim}\limits_{h \to 0}\frac{-10x-5h}{\left(1+x^{2}+2xh+h^{2}\right)\left(1+x^{2}\right)}=\]\[\frac{-10x}{\left(1+x^{2}\right)^{2}}\]
- Substitute the \(x\)-coordinate into the derivative to find \(m\) \[m=f'(2)=\frac{-10\left(2\right)}{\left(1+\left(2\right)^{2}\right)^{2}}=\frac{-20}{25}-\frac{4}{5}\]
- Find the \(y\)-coordinate \[f(2)=\frac{5}{1+4}=1\]
- Write the equation of the tangent line at the point \[y-1=-\frac{4}{5}\left(x-2\right)\]\[y=y=-\frac{4}{5}x+\frac{13}{5}=\frac{13-4x}{5}\]
EX 6
Using the alternate form of the derivative find \(f'(2)\) of \(\sqrt{x+2}\).
Using the formula:
\[f'\left(2\right)=\mathop {\lim}\limits_{x \to 2}\frac{\sqrt{x+2}-2}{x-2}=\]\[\mathop {\lim}\limits_{x \to 2}\frac{\left(\sqrt{x+2}-2\right)\left(\sqrt{x+2}+2\right)}{\left(x-2\right)\left(\sqrt{x+2}+2\right)}=\mathop {\lim}\limits_{x \to 2}\frac{x+2-4}{\left(x-2\right)\left(\sqrt{x+2}+2\right)}=\]\[\mathop {\lim}\limits_{x \to 2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+2}+2\right)}=\mathop {\lim}\limits_{x \to 2}\frac{x-2}{\left(x-2\right)\left(\sqrt{x+2}+2\right)}=\]\[\mathop {\lim}\limits_{x \to 2}\frac{1}{\left(\sqrt{x+2}+2\right)}=\frac{1}{2}\]
EX 7
Find the equation of the tangent line to the parabola \(y=3x^{2}-2x+9\) at the point \(x=2\) using the alternate form of the derivative.
- Find \(m\) using the formula \[f'\left(2\right)=\mathop {\lim}\limits_{x \to 2}\frac{3x^{2}-2x+9-3\left(2\right)^{2}+2\left(2\right)-9}{x-2}=\]\[\mathop {\lim}\limits_{x \to 2}\frac{3x^{2}-2x-8}{x-2}=\mathop {\lim}\limits_{x \to 2}\frac{\left(x-2\right)\left(3x+4\right)}{x-2}=\]\[\mathop {\lim}\limits_{x \to 2}\left(3x+4\right)=10\] \[\therefore\] \[m=10\]
- Find the \(y\)-coordinate \[f\left(2\right)=3\left(2\right)^{2}-2\left(2\right)+9=17\]
- Write the equation of the tangent line at the point \[y-17=10\left(x-2\right)\] \[y=10x-20+17=10x-3\]
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