Last modified on 2 November 2011, at 01:39

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}