Introduction

Welcome! In this article, we will not be learning a specific concept, but rather reviewing a skill that is very important in Calculus. We have introduced many rules and gone over them in great depth, but knowing when to apply each rule is something we haven’t discussed that is worth having an article on.

Up to this point you have learned many different rules and procedures for calculating derivatives, the point of this topic is to test your knowledge of what you have learned prior. In this article we will go over examples of differentiating difficult functions with everything you have learned so far about derivatives. For each of these example problems I would like you to attempt to differentiate it yourself first before looking over what I do.

Problem #1

Find the derivative of $f(x)=\frac{\ln ([\arcsin (x){]}^2)+\ln (e^{2x})}{{\log }_7(x^2-x)-{\log }_7(x)}$

This is a very complex function, and will require many different derivative rules working together. The first step we can do is to try to simplify this one by one using logarithmic properties. 

The numerator can be simplified like this: 

$\ln ([\arcsin (x){]}^2)+\ln (e^{2x})=2\ln (\arcsin (x))+2x$

For the denominator, we can do this:

${\log }_7(x^2-x)-{\log }_7(x)={\log }_7\left(\frac{x^2-x}{x}\right)={\log }_7(x-1)$

Our new simplified function is:

$\frac{2\ln (\arcsin (x))+2x}{{\log }_7(x-1)}$

Notice that we have a quotient, so we will have to apply the quotient rule. We will have to find the derivative of both the numerator and denominator to apply the quotient rule. 

$f(x)=2\ln (\arcsin (x))+2x$

$g(x)={\log }_7(x-1)$

To solve for $f’(x)$, we need to use the chain rule on the $2\ln (\arcsin (x))$ term. The outer function is $2\ln (x)$ and the inner function is $\arcsin (x)$. The derivative of the outer function is $\frac{2}{x}$, and the derivative of the inner function is $\frac{1}{\sqrt{1-x^2}}$. If $f’(x)=\frac{2}{x}$, then $f’(\arcsin (x))=\frac{2}{\arcsin (x)}$. The derivative of $2\ln (\arcsin (x))$ is $\frac{2}{\arcsin (x)}\cdot \frac{1}{\sqrt{1-x^2}}=\frac{2}{\arcsin (x))\sqrt{1-x^2}}$

To solve for $g’(x)$, we can simply use the derivative for logarithmic rules to differentiate ${\log }_7(x-1)$. The derivative is $\frac{1}{(x-1)\ln 7}$.

Now that we have found the derivatives of both the numerator and denominator, we can simply put everything together to get the final answer.

$\frac{(\frac{2}{\arcsin (x))\sqrt{1-x^2}})({\log }_7(x-1))-(\frac{1}{(x-1)\ln 7})(2\ln (\arcsin (x))+2x)}{[{\log }_7(x-1){]}^2}$

What an extremely tedious problem. Hopefully you were able to follow along. Notice that we simplified the problem to make it easier to differentiate, and used many different rules.

Problem #2

Lets try another one, find the derivative, $\frac{dy}{dx}$, of the following function:

$y=-5\tan (\arccos (\ln (e^{3x})))$

First simply the function: $y=-5\tan (\arccos (3x))$

Then just solve by applying the chain rule:

$\frac{dy}{dx}=-5{\sec }^2(\arccos (3x))\cdot -\frac{1}{\sqrt{1-(3x{)}^2}}\cdot 3$

$\frac{dy}{dx}=\frac{15{\sec }^2(\arccos (3x))}{\sqrt{1-(3x{)}^2}}$

$\frac{dy}{dx}=\frac{15{\sec }^2(\arccos (3x))}{\sqrt{1-9x^2}}$

Something else you might see when trying to simplify a function before solving it is a logarithm with the same base as an exponential function. In this situation we had $\ln (x)$ and $e^x$ and since both are inverse functions we could cancel them out before solving.

Problem #3

This one will test to see if you really know your trig derivatives, find the derivative, $f’$, of the following function:

$f(x)=\sec (\cos (\tan (x)))$

Here all you need to do is apply the chain rule and use your knowledge of trig derivatives:$\frac{d}{dx}[\sec (x)]=\sec (x)\tan (x)$

$\frac{d}{dx}[\cos (x)]=-\sin (x)$

$\frac{d}{dx}[\tan (x)]={\sec }^2(x)$

$f’(x)=\sec (\cos (\tan (x)))\cdot \tan (\cos (\tan (x)))\cdot -\sin (\tan (x))\cdot {\sec }^2(x)$

$=-{\sec }^2(x)\sin (\tan (x))\sec (\cos (\tan (x)))\tan (\cos (\tan (x)))$

As long as you know all your trig derivatives problems like this should be a breeze. Things do start to get messy after multiple applications of the chain rule though. While derivative problems like the ones in this article might seem a bit more difficult than you are used to, all of them should be relatively straightforward to solve as long as you know all your derivative rules.

Problem #4

Solve for the derivative, $\frac{dy}{dx}$, of the following equation:

$\frac{5}{\sqrt{y}}+\frac{x^2}{y}-{\cos }^2(5x)=27$

First convert the fractions into exponents so we can apply the power rule:

$5y^{-\frac{1}{2}}+x^2{y}^{-1}-{\cos }^2(5x)=27$

Then just differentiate both sides with respect to $x$ and solve for $\frac{dy}{dx}$:

$\frac{d}{dx}[5y^{-\frac{1}{2}}+x^2{y}^{-1}-{\cos }^2(5x)]=\frac{d}{dx}[27]$

$[-\frac{5}{2}{y}^{-\frac{3}{2}}(\frac{dy}{dx})]+[2x{y}^{-1}-x^2{y}^{-2}(\frac{dy}{dx})]-[2\cos (5x)\cdot -\sin (5x)\cdot 5]=0$

$-\frac{5}{2}{y}^{-\frac{3}{2}}(\frac{dy}{dx})+2x{y}^{-1}-x^2{y}^{-2}(\frac{dy}{dx})+10\cos (5x)\sin (5x)=0$

$-\frac{5}{2}{y}^{-\frac{3}{2}}(\frac{dy}{dx})-x^2{y}^{-2}(\frac{dy}{dx})=-2x{y}^{-1}-10\cos (5x)\sin (5x)$

$\frac{5}{2}{y}^{-\frac{3}{2}}(\frac{dy}{dx})+x^2{y}^{-2}(\frac{dy}{dx})=2x{y}^{-1}+10\cos (5x)\sin (5x)$

$\frac{dy}{dx}=\frac{2x{y}^{-1}+10\cos (5x)\sin (5x)}{\frac{5}{2}{y}^{-\frac{3}{2}}+x^2{y}^{-2}}$

The actual differentiation used in this problem is quite simple compared to the previous problems which have just gone over. What makes this one seem so hard is the algebraic manipulation that has to be used to reach the answer.

Conclusion and Notes on Derivatives.

What should be the hardest part of these problems is the algebra. Again as long as you know all of the basic rules for differentiation and derivatives of common functions, solving these problems should be relatively straightforward. If you found any of the concepts used in these problems difficult I recommend reviewing them. From this point onwards, though you might not expand to much more on derivatives themselves, you will learn many ways to apply them. For those applications it’s important that you understand the fundamentals well.

The Rules of Differentiation.

I would like to provide a comprehensive list of all the derivative rules you should know Units 2 and 3, as well as a list of all the derivatives of common functions.

Rules:

Constant Rule: $\frac{d}{dx}(C)=0$

Constant Multiple Rule: $\frac{d}{dx}[C\cdot f(x)]=C\cdot f’(x)$

Power Rule: $\frac{d}{dx}(x^n)=n{x}^{n-1}$

Sum Rule: $\frac{d}{dx}[f(x)+g(x)]=f’(x)+g’(x)$

Difference Rule: $\frac{d}{dx}[f(x)-g(x)]=f’(x)-g’(x)$

Product Rule: $\frac{d}{dx}[f(x)\cdot g(x)]=f’(x)g(x)+f(x)g’(x)$

Quotient Rule: $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]=\frac{f’(x)g(x)-f(x)g’(x)}{[g(x){]}^2}$

Chain Rule: $\frac{d}{dx}[f(g(x))]=f’(g(x))\cdot g’(x)$

Inverse Function Rule: $\frac{d}{dx}[f^{-1}(x)]=\frac{1}{f’(f^{-1}(x))}$

Common Function Derivatives:

$\frac{d}{dx}[\sin (x)]=\cos (x)$

$\frac{d}{dx}[\cos (x)]=-\sin (x)$

$\frac{d}{dx}[\tan (x)]={\sec }^2(x)$

$\frac{d}{dx}[\sec (x)]=\sec (x)\tan (x)$

$\frac{d}{dx}[\csc (x)]=-\csc (x)\cot (x)$

$\frac{d}{dx}[\cot (x)]=-cs{c}^2(x)$

$\frac{d}{dx}[\arcsin (x)]=\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\arccos (x)]=-\frac{1}{\sqrt{1-x^2}}$

$\frac{d}{dx}[\arctan (x)]=\frac{1}{1+x^2}$

$\frac{d}{dx}[{\cot }^{-1}(x)]=-\frac{1}{1+x^2}$

$\frac{d}{dx}[e^x]=e^x$

$\frac{d}{dx}[\ln (x)]=\frac{1}{x}$

$\frac{d}{dx}[a^x]=a^x\ln (a)$

$\frac{d}{dx}[{\log }_a(x)]=\frac{1}{{\log }_a(x)}$