# Calculus/Differentiation/Differentiation Defined/Solutions

1. Find the slope of the tangent to the curve $y=x^{2}$ at $(1,1)$ .
The definition of the slope of $f$ at $x_{0}$ is $\lim _{h\to 0}\left[{\frac {f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}}\right]$ Substituting in $f(x)=x^{2}$ and $x_{0}=1$ gives:

{\begin{aligned}\lim _{h\to 0}\left[{\frac {(1+h)^{2}-1}{h}}\right]&=\lim _{h\to 0}\left[{\frac {h^{2}+2h}{h}}\right]\\&=\lim _{h\to 0}\left[{\frac {h(h+2)}{h}}\right]\\&=\lim _{h\to 0}h+2\\&=\mathbf {2} \end{aligned}} The definition of the slope of $f$ at $x_{0}$ is $\lim _{h\to 0}\left[{\frac {f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h}}\right]$ Substituting in $f(x)=x^{2}$ and $x_{0}=1$ gives:

{\begin{aligned}\lim _{h\to 0}\left[{\frac {(1+h)^{2}-1}{h}}\right]&=\lim _{h\to 0}\left[{\frac {h^{2}+2h}{h}}\right]\\&=\lim _{h\to 0}\left[{\frac {h(h+2)}{h}}\right]\\&=\lim _{h\to 0}h+2\\&=\mathbf {2} \end{aligned}} 2. Using the definition of the derivative find the derivative of the function $f(x)=2x+3$ .
{\begin{aligned}f^{'}(x)&=\lim _{\Delta x\to 0}{\frac {(2(x+\Delta x)+3)-(2x+3)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x+2\Delta x+3-2x-3)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}2\\&=\mathbf {2} \end{aligned}} {\begin{aligned}f^{'}(x)&=\lim _{\Delta x\to 0}{\frac {(2(x+\Delta x)+3)-(2x+3)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2x+2\Delta x+3-2x-3)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {2\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0}2\\&=\mathbf {2} \end{aligned}} 3. Using the definition of the derivative find the derivative of the function $f(x)=x^{3}$ . Now try $f(x)=x^{4}$ . Can you see a pattern? In the next section we will find the derivative of $f(x)=x^{n}$ for all $n$ .
{\begin{alignedat}{2}{\frac {dx^{3}}{dx}}&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{3}-x^{3}}{\Delta x}}&\qquad {\frac {dx^{4}}{dx}}&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{4}-x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {x^{3}+3x^{2}\Delta x+3x\Delta x^{2}+\Delta x^{3}-x^{3}}{\Delta x}}&&=\lim _{\Delta x\to 0}{\frac {x^{4}+4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4}-x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {3x^{2}\Delta x+3x\Delta x^{2}+\Delta x^{3}}{\Delta x}}&&=\lim _{\Delta x\to 0}{\frac {4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}3x^{2}+3x\Delta x+\Delta x^{2}&&=\lim _{\Delta x\to 0}4x^{3}+6x^{2}\Delta x+4x\Delta x^{2}+\Delta x^{3}\\&=\mathbf {3x^{2}} &&=\mathbf {4x^{3}} \end{alignedat}} {\begin{alignedat}{2}{\frac {dx^{3}}{dx}}&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{3}-x^{3}}{\Delta x}}&\qquad {\frac {dx^{4}}{dx}}&=\lim _{\Delta x\to 0}{\frac {(x+\Delta x)^{4}-x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {x^{3}+3x^{2}\Delta x+3x\Delta x^{2}+\Delta x^{3}-x^{3}}{\Delta x}}&&=\lim _{\Delta x\to 0}{\frac {x^{4}+4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4}-x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {3x^{2}\Delta x+3x\Delta x^{2}+\Delta x^{3}}{\Delta x}}&&=\lim _{\Delta x\to 0}{\frac {4x^{3}\Delta x+6x^{2}\Delta x^{2}+4x\Delta x^{3}+\Delta x^{4}}{\Delta x}}\\&=\lim _{\Delta x\to 0}3x^{2}+3x\Delta x+\Delta x^{2}&&=\lim _{\Delta x\to 0}4x^{3}+6x^{2}\Delta x+4x\Delta x^{2}+\Delta x^{3}\\&=\mathbf {3x^{2}} &&=\mathbf {4x^{3}} \end{alignedat}} 4. The text states that the derivative of $\left|x\right|$ is not defined at $x=0$ . Use the definition of the derivative to show this.
{\begin{alignedat}{2}\lim _{\Delta x\to 0^{-}}{\frac {\left|0+\Delta x\right|-\left|0\right|}{\Delta x}}&=\lim _{\Delta x\to 0^{-}}{\frac {-\Delta x}{\Delta x}}&\qquad \lim _{\Delta x\to 0^{+}}{\frac {\left|0+\Delta x\right|-\left|0\right|}{\Delta x}}&=\lim _{\Delta x\to 0^{+}}{\frac {\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0^{-}}-1&&=\lim _{\Delta x\to 0^{+}}1\\&=-1&&=1\end{alignedat}} Since the limits from the left and the right at $x=0$ are not equal, the limit does not exist, so $\left|x\right|$ is not differentiable at $x=0$ .
{\begin{alignedat}{2}\lim _{\Delta x\to 0^{-}}{\frac {\left|0+\Delta x\right|-\left|0\right|}{\Delta x}}&=\lim _{\Delta x\to 0^{-}}{\frac {-\Delta x}{\Delta x}}&\qquad \lim _{\Delta x\to 0^{+}}{\frac {\left|0+\Delta x\right|-\left|0\right|}{\Delta x}}&=\lim _{\Delta x\to 0^{+}}{\frac {\Delta x}{\Delta x}}\\&=\lim _{\Delta x\to 0^{-}}-1&&=\lim _{\Delta x\to 0^{+}}1\\&=-1&&=1\end{alignedat}} Since the limits from the left and the right at $x=0$ are not equal, the limit does not exist, so $\left|x\right|$ is not differentiable at $x=0$ .
6. Use the definition of the derivative to show that the derivative of $\sin x$ is $\cos x$ . Hint: Use a suitable sum to product formula and the fact that $\lim _{t\to 0}{\frac {\sin(t)}{t}}=1$ and $\lim _{t\to 0}{\frac {\cos(t)-1}{t}}=0$ .
{\begin{aligned}\lim _{\Delta x\to 0}{\frac {\sin(x+\Delta x)-\sin(x)}{\Delta x}}&=\lim _{\Delta x\to 0}{\frac {(\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x))-\sin(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\sin(x)(\cos(\Delta x)-1)+\cos(x)\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot \lim _{\Delta x\to 0}{\frac {\cos(\Delta x)-1}{\Delta x}}+\cos(x)\cdot \lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot 0+\cos(x)\cdot 1\\&=\cos(x)\end{aligned}} {\begin{aligned}\lim _{\Delta x\to 0}{\frac {\sin(x+\Delta x)-\sin(x)}{\Delta x}}&=\lim _{\Delta x\to 0}{\frac {(\sin(x)\cos(\Delta x)+\cos(x)\sin(\Delta x))-\sin(x)}{\Delta x}}\\&=\lim _{\Delta x\to 0}{\frac {\sin(x)(\cos(\Delta x)-1)+\cos(x)\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot \lim _{\Delta x\to 0}{\frac {\cos(\Delta x)-1}{\Delta x}}+\cos(x)\cdot \lim _{\Delta x\to 0}{\frac {\sin(\Delta x)}{\Delta x}}\\&=\sin(x)\cdot 0+\cos(x)\cdot 1\\&=\cos(x)\end{aligned}} • Find the derivatives of the following equations:
7. $f(x)=42$ $\mathbf {f'(x)=0}$ $\mathbf {f'(x)=0}$ 8. $f(x)=6x+10$ $\mathbf {f'(x)=6}$ $\mathbf {f'(x)=6}$ 9. $f(x)=2x^{2}+12x+3$ $\mathbf {f'(x)=4x+12}$ $\mathbf {f'(x)=4x+12}$ 