Derivative of Polynomials and Trigonometric Functions - Practice Questions & MCQ

Derivative of the Polynomial Function

Finding derivatives of functions by using the definition of the derivative (by using the First Principle) can be a lengthy and, for certain functions, a rather challenging process. In this section, we will learn direct results for finding derivatives of certain standard functions that allow us to bypass this process.

$\begin{array}{rl}& \text{ 1. }\frac{d}{dx}(\text{ constant })=\mathbf{0}\\ & \begin{array}{rl}f(x)& =c\text{ and }f(x+h)=c\\ f^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{c-c}{h}\\ & =\underset{h\to 0}{lim}\frac{0}{h}\\ & =\underset{h\to 0}{lim}0=0\end{array}\end{array}$
So the derivative of a constant function is zero. Also since a constant function is a horizontal line the slope of a constant function is 0.
2. $\frac{d}{dx}\left({\mathbf{x}}^{\mathrm{n}}\right)=\mathbf{n}{\mathbf{x}}^{\mathrm{n}-1}$

For $f(x)=x^n$ where $n$ is a positive integer, we have

$f^{\prime }(x)=\underset{h\to 0}{lim}\frac{(x+h{)}^n-x^n}{h}$
Since,

$(x+h{)}^n=x^n+n{x}^{n-1}h+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}{h}^2+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}{h}^3+\dots +nx{h}^{n-1}+h^n$

We can write

$(x+h{)}^n-x^n=n{x}^{n-1}h+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}{h}^2+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}{h}^3+\dots +nx{h}^{n-1}+h^n$
Next, divide both sides by h:

$\frac{(x+h{)}^n-x^n}{h}=\frac{n{x}^{n-1}h+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}{h}^2+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}{h}^3+\dots +nx{h}^{n-1}+h^n}{h}$
Thus,

$\frac{(x+h{)}^n-x^n}{h}=n{x}^{n-1}+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}h+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}{h}^2+\dots +nx{h}^{n-2}+h^{n-1}$
Finally,

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{(x+h{)}^n-x^n}{h}\\ & =\underset{h\to 0}{lim}(n{x}^{n-1}+(\genfrac{}{}{0ex}{}{n}{2})x^{n-2}h+(\genfrac{}{}{0ex}{}{n}{3})x^{n-3}{h}^2+\dots +nx{h}^{n-1}+h^n)\\ & =n{x}^{n-1}\end{array}$

Derivative of Exponential and Logarithmic Functions

4. $\frac{d}{dx}\left({\mathbf{a}}^{\mathbf{x}}\right)={\mathbf{a}}^{\mathbf{x}}{\log }_{\mathrm{e}}\mathbf{a}$
$f(x)=a^x$ and $f(x+h)=a^{x+h}$

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{a^{x+h}-a^x}{h}\\ & =a^x\underset{h\to 0}{lim}\frac{a^h-1}{h}\\ & =a^x{\log }_{e}a& [\because \underset{x\to 0}{lim}\frac{a^x-1}{x}={\log }_{e}a]\end{array}$

5. $\frac{d}{dx}\left({\mathbf{e}}^{\mathbf{x}}\right)={\mathbf{e}}^{\mathbf{x}}{\log }_{\mathbf{e}}\mathbf{e}={\mathbf{e}}^{\mathbf{x}}$

6. $\frac{d}{dx}({\log }_{\mathrm{e}}|\mathbf{x}|)=\frac{\mathbf{1}}{\mathbf{x}},\mathbf{x}\ne 0$
$f(x)={\log }_e(x)$ and $f(x+h)={\log }_e(x+h)$

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{{\log }_e(x+h)-{\log }_e(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{{\log }_e\left(\frac{x+h}{x}\right)}{h}\\ & =\underset{h\to 0}{lim}\frac{{\log }_e(1+\frac{h}{x})}{\frac{h}{x}\cdot x}\\ & =\frac{1}{x}[\because \underset{x\to 0}{lim}\frac{{\log }_e(1+x)}{x}=1]\end{array}$

7. $\frac{d}{dx}({\log }_{\mathbf{a}}|\mathbf{x}|)=\frac{1}{\mathbf{x}{\log }_{\mathbf{e}}\mathbf{a}},\mathbf{x}\ne \mathbf{0}$

Derivative of the Trigonometric Functions (sin/cos/tan)

8. $\frac{d}{dx}(\sin (\mathbf{x}))=\cos (\mathbf{x})$
$f(x)=\sin (x)$ and $f(x+h)=\sin (x+h)$

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{lim}\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{lim}\frac{\sin (x+h)-\sin (x)}{h}\\ & =\underset{h\to 0}{lim}\frac{2\sin \frac{h}{2}\cdot \cos \left(\frac{2x+h}{2}\right)}{h}\\ & =\underset{h\to 0}{lim}\frac{\sin \frac{h}{2}}{\frac{h}{2}}\cdot \underset{h\to 0}{lim}\cos \left(\frac{2x+h}{2}\right)\\ & =\cos x[\because \underset{x\to 0}{lim}\frac{\sin (x)}{x}=1]\end{array}$

9. $\frac{d}{dx}(\cos (\mathbf{x}))=-\sin (\mathbf{x})$
10. $\frac{d}{dx}(\tan (\mathbf{x}))={\sec }^2(\mathbf{x})$

Derivative of the Trigonometric Functions (cot/sec/csc)

11. $\frac{d}{dx}(\cot (\mathrm{x}))=-{\csc }^2(\mathrm{x})$
12. $\frac{d}{dx}(\sec (\mathbf{x}))=\sec (\mathbf{x})\tan (\mathbf{x})$
13. $\frac{d}{dx}(\csc (\mathbf{x}))=-\csc (\mathbf{x})\cot (\mathbf{x})$