# 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{align} \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{align}

2. Using the definition of the derivative find the derivative of the function $f(x)=2x+3$.

\begin{align}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{align}

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{alignat}{2}\frac{d x^3}{dx} &=\lim_{\Delta x\to 0}\frac{(x+\Delta x)^3-x^3}{\Delta x} & \qquad\frac{d x^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{alignat}

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{alignat}{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{alignat}
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{align}\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{align}

• Find the derivatives of the following equations:
7. $f(x) = 42$

$\mathbf{f'(x)=0}$

8. $f(x) = 6x + 10$

$\mathbf{f'(x)=6}$

9. $f(x) = 2x^2 + 12x + 3$

$\mathbf{f'(x)=4x+12}$