## Exercises - Basic Derivative Rules (and Review)

1. Find the derivative of each function given

1. $\displaystyle{f\,(x) = 4}$

2. $\displaystyle{f\,(x) = e^{\pi}}$

3. $\displaystyle{f\,(x) = \sqrt{2}}$

$0$ for (a), (b), and (c)     [Each is a constant function.]

2. Find $f\,'(x)$ for each function below

1. $\displaystyle{f\,(x)=x^4}$

2. $\displaystyle{f\,(x)=x^{100}}$

3. $\displaystyle{f\,(x)=4x^3}$

4. $\displaystyle{f\,(x)=\frac{1}{2} x^{2/3}}$

5. $\displaystyle{f\,(x)=4x^{-1/4}}$

6. $\displaystyle{f\,(x)=\pi x^{\sqrt{2}}}$

We apply the Power Rule for derivatives in each case:

1. $4x^3$
2. $100x^{99}$
3. $12x^2$
4. $\displaystyle{\frac{1}{3x^{1/3}}}$
5. $\displaystyle{-\frac{1}{x^{5/4}}}$
6. $\pi\sqrt{2} \cdot x^{\sqrt{2}-1}$
3. In each case, find $y'$

1. $\displaystyle{y = 4\sqrt{x}+\frac{3}{x}-5x+4}$

2. $\displaystyle{y = \frac{2}{\sqrt{x}} + \frac{\sqrt{x}}{2} + \frac{1}{x} - x}$

3. $\displaystyle{y = \sqrt{x} + \frac{1}{\sqrt{x}} + \frac{3}{x} - 3x}$

1. $\displaystyle{\frac{2}{\sqrt{x}} - \frac{3}{x^2} - 5}$

2. $\displaystyle{-\frac{1}{x^{3/2}} + \frac{1}{4\sqrt{x}} - \frac{1}{x^2} - 1}$

3. $\displaystyle{\frac{1}{2\sqrt{x}} - \frac{1}{2x^{3/2}} - \frac{3}{x^2} - 3}$

4. Find $\displaystyle{\frac{d}{dx}(4x^3 - 3x^2 + 2)}$.

$12x^2-6x$

5. Differentiate, simplifying your result as appropriate

1. $\displaystyle{f\,(x) = 7x^3+6x^2+10x+12}$

2. $\displaystyle{f\,(x) = 3x^3-6x^{-2}+x^{-7}-x^{-6}}$

1. $21x^2 + 12x + 10$

2. $\displaystyle{9x^2 + \frac{12}{x^3} - \frac{7}{x^8} + \frac{6}{x^7}}$

6. Find the equation of the tangent line to $y=x^3+3x^2+3x+9$ where $x=-3$

$y' = 3x^2 + 6x + 3$, so when $x = -3$, slope $m = y' = 27 -18 + 3 = 12$ and $y = -27 + 27 - 9 + 9 = 0$. Using point-slope form, we then have $y = 12(x+3)$.

7. In each case, find $y'$

1. $\displaystyle{y = (x^2 - \frac{1}{x})(3x-4)}$

2. $\displaystyle{y = (\sqrt{x} + 5)(x^2 - 5)}$

3. $\displaystyle{y = (\sin x)(\cos x) - x \cos x + x^3}$

4. $\displaystyle{y = x^3 \sin x - x^3 \cos x}$

5. $\displaystyle{y = \ln e^x - e^x \ln x}$

1. $\displaystyle{9x^2 - 8x - \frac{4}{x^2}}$

2. $\displaystyle{2x\sqrt{x} + 10x + \frac{x\sqrt{x}}{2} - \frac{5}{2\sqrt{x}}}$

3. $\displaystyle{-\sin^2 x + \cos^2 x + x\sin x - \cos x + 3x^2}$

4. $\displaystyle{x^3 \cos x + x^3 \sin x + 3x^2 \sin x - 3x^2 \cos x}$

5. $\displaystyle{1 - \frac{e^x}{x} - e^x \ln x}$

8. Differentiate:

1. $\displaystyle{f\,(x) = (x^2+x)(3x+1)}$

2. $\displaystyle{f\,(x) = x^2e^x + \frac{1}{x^2}}$

3. $\displaystyle{f\,(x) = x^3\sin x + \sin (\frac{\pi}{4} )}$

1. $\displaystyle{9x^2+8x+1}$

2. $\displaystyle{x^2 e^x + 2x e^x - \frac{2}{x^3}}$

3. $\displaystyle{x^3 \cos x + 3 x^2 \sin x}$

9. Differentiate the following

1. $\displaystyle{y = \frac{\sin x}{\ln x}}$

2. $\displaystyle{y = \frac{x}{x-4}}$

3. $\displaystyle{y = \frac{x^2 - 1}{(x+4)^2}}$

1. $\displaystyle{y' = \frac{x (\ln x)(\cos x) - \sin x}{x \ln^2 x}}$

2. $\displaystyle{y' = \frac{-4}{(x-4)^2}}$

3. $\displaystyle{y' = \frac{2(4x+1)}{(x+4)^3}}$

10. Suppose $\displaystyle{y = \frac{x^2-4x+3}{x-1}}$ and then find $\displaystyle{\frac{dy}{dx}}$

We can work this problem in two ways. One way uses the quotient rule. Another realizes that this function can be made more "differentiation-friendly" by simplifying first. Just make sure you realize the implications with regard to the domain if you take the second, much easier, way:

$$y = \frac{x^2-4x+3}{x-1} = \frac{(x-1)(x-3)}{x-1} = x-3 \quad \textrm{for all } x \ne 1$$

Thus, $\displaystyle{\frac{dy}{dx} = \frac{d}{dx} (x-3) = 1}$ when $x \ne 1$, and is undefined if $x=1$.

11. Show that if $f$ is differentiable, then use the definition of the derivative to show that $\displaystyle{\frac{d}{dx} \left[ f(x) \right]^2 = 2 f(x) f'(x)}$. Is this consistent with the basic derivative rules?

$$\begin{array}{rcl} \displaystyle{\frac{d}{dx} \left[ f(x) \right]^2} &=& \displaystyle{\lim_{h \rightarrow 0} \frac{\left[ f(x+h) \right]^2 - \left[ f(x) \right]^2}{h}}\\ &=& \displaystyle{\frac{1}{h} \left( [f(x+h) - f(x)][f(x+h) + f(x)] \right)}\\ &=& \displaystyle{\left( \lim_{h \rightarrow 0} [f(x+h) + f(x)] \right) \cdot \left( \lim_{h \rightarrow 0} \frac{f(x+h) - f(x)}{h} \right)}\\ &=& \displaystyle{2f(x) f'(x)} \end{array}$$

The basic derivative rules lead (more quickly) to the same conclusion. Consider either a direct application of the product rule on $f(x) \cdot f(x)$, or the chain rule with an inner function of $f(x)$ and an outer function of $x^2$.

12. Show that if $f$ is differentiable everywhere and $g(x) = f(x^2)$, then $g'(c) = 2c f'(c^2)$. Is this consistent with the basic derivative rules?

Noting that $\displaystyle{f'(c^2) = \lim_{u \rightarrow c^2} \frac{f(u) - f(c^2)}{u-c^2}}$ guides our work...

$$\begin{array}{rcl} g'(c) &=& \displaystyle{ \lim_{x \rightarrow c} \frac{g(x) - g(c)}{x-c}}\\\\ &=& \displaystyle{ \lim_{x \rightarrow c} \frac{f(x^2) - f(c^2)}{x-c} \quad \textrm{we'll need } u = x^2 \textrm{ and then } u-c^2 = x^2 - c^2 \textrm{ in the bottom}}\\\\ &=& \displaystyle{ \lim_{x \rightarrow c} \frac{f(x^2) - f(c^2)}{x-c} \cdot \frac{x^2-c^2}{x^2-c^2}}\\\\ &=& \displaystyle{ \left( \lim_{x \rightarrow c} \frac{f(x^2) - f(c^2)}{x^2-c^2} \right) \cdot \left( \lim_{x \rightarrow c} \frac{x^2-c^2}{x-c} \right) \quad \textrm{now write the limit in terms of } u}\\\\ &=& \displaystyle{ \left( \lim_{u \rightarrow c^2} \frac{f(u) - f(c^2)}{u-c^2} \right) \cdot \left( \lim_{x \rightarrow c} \frac{x^2-c^2}{x-c} \right)}\\\\ &=& \displaystyle{ \left( \lim_{u \rightarrow c^2} \frac{f(u) - f(c^2)}{u-c^2} \right) \cdot \left( \lim_{x \rightarrow c} \, (x+c) \right)}\\\\ &=& (f'(c^2))(2c) \end{array}$$

The basic derivative rules lead (more quickly) to the same conclusion. Consider a direct application of the Chain Rule to $g(x)$ where the inner functions is $x^2$ and the outer function is $f(x)$.

13. Instead of the binomial theorem, use induction and the product rule to show that $\displaystyle{\frac{d}{dx} x^n = n x^{n-1}}$ for positive integers $n$.

First we show the base case holds using the limit definition of the derivative:

$$\frac{d}{dx} x = \lim_{h \rightarrow 0} \frac{(x+h)-x}{h} = \lim_{h \rightarrow 0} \frac{h}{h} = 1$$

Then we assume that for some $k$, we know $\displaystyle{\frac{d}{dx} x^k = k x^{k-1}}$, and show with that assumption (and the product rule) it must be the case that: $$\displaystyle{\frac{d}{dx} x^{k+1} = \frac{d}{dx} [x \cdot x^k] = x \cdot kx^{k-1} + x^k \cdot 1 = k x^k + x^k = (k+1) x^k}$$

Hence, by induction the desired result holds.

14. (Challenge Problem) Use $\displaystyle{e = \lim_{x \rightarrow \infty} \left(1 + \frac{1}{n} \right)^n}$ to prove $\displaystyle{\frac{d}{dx} e^x} = e^x$.

$$\begin{array}{rcl} \displaystyle{\frac{d}{dx} e^x} &=& \displaystyle{\lim_{h \rightarrow 0} \frac{e^{x+h} - e^x}{h}}\\\\ &=& \displaystyle{\lim_{h \rightarrow 0} e^x \cdot \left( \frac{e^h - 1}{h} \right)}\\\\ &=& \displaystyle{e^x \cdot \lim_{h \rightarrow 0} \frac{e^h - 1}{h} \quad \textrm{Now let } u = e^h - 1. \textrm{ Thus, } \lim_{h \rightarrow 0} u = 0 \textrm{ and } h = \ln(u+1)}\\\\ &=& \displaystyle{e^x \cdot \lim_{u \rightarrow 0} \frac{u}{\ln(u+1)}}\\\\ &=& \displaystyle{e^x \cdot \lim_{u \rightarrow 0} \left( \frac{\ln \, (u+1)}{u} \right)^{-1}}\\\\ &=& \displaystyle{e^x \cdot \lim_{u \rightarrow 0} \left( \frac{1}{u} \ln \, (u+1) \right)^{-1}}\\\\ &=& \displaystyle{e^x \cdot \lim_{u \rightarrow 0^+} \left( \ln \, (u+1)^{1/u} \right)^{-1}}\\\\ &=& \displaystyle{e^x \cdot \lim_{u \rightarrow 0^+} \left( \ln \, (u+1)^{1/u} \right)^{-1} \quad \textrm{Now let } w = \frac{1}{u} \textrm{ and note } \lim_{u \rightarrow 0^+} w = \infty}\\\\ &=& \displaystyle{e^x \cdot \lim_{w \rightarrow \infty} \left[ \ln \, \left(\frac{1}{w} + 1\right)^w \right]^{-1}}\\\\ &=& \displaystyle{e^x \cdot \left\{ \ln \left[ \lim_{w \rightarrow \infty} \left(1 + \frac{1}{w}\right)^w \right] \right\}^{-1}}\\\\ &=& \displaystyle{e^x \cdot (\ln e)^{-1}} \quad \textrm{ ...after applying the definition for } e \textrm{ given in the problem.}\\\\ &=& e^x \cdot 1\\\\ &=& e^x \end{array}$$