Calculus/Differentiation/Differentiation Defined/Solutions

< Calculus < Differentiation < Differentiation Defined

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}}

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}}

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}}

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

.

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}}

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}

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