55 816
Assignments Done
97,2%
Successfully Done
In December 2017
Your physics homework can be a real challenge, and the due date can be really close — feel free to use our assistance and get the desired result.
Be sure that math assignments completed by our experts will be error-free and done according to your instructions specified in the submitted order form.
Our experts will gladly share their knowledge and help you with programming homework. Keep up with the world’s newest programming trends.
Get a free quote.
Check the website
for updates.
Receive your completed assignment.
Easy as ABC!
Just provide us with clear instructions and wait for the completed assignment.

How to Apply Chain Rule and Quotient Rule Simultaneously?

Chain rule and Quotient Rule Simultaneously

Previously we’ve discussed the chain rule for derivatives and how to apply it. However, sometimes using chain rule alone is not enough and you’re also required to apply other rules for derivatives to get the correct answer. Today we’ll consider one of such cases. Let’s find derivative of the following function:

$$h(x)=\frac{\sin^5{(3x+1) }}{(4x+1)^3}$$

Notice that given function can be represented as a ratio of two functions @$\frac{f(x)}{g(x)}@$ where

$$f(x)=\sin^5{(3x+1) }$$

$$g(x)=(4x+1)^3$$

As we have quotient, this means we need to use quotient rule for derivatives. Since @$f(x), g(x)@$ are both composite functions, for each of them we should apply chain rule when finding derivatives.

Recall quotient rule:

$$h'(x)=(\frac{f(x)}{g(x)} )’=\frac{f’ (x)g(x)-f(x)g'(x)}{g(x)^2}$$

First let’s find derivatives @$f’ (x),g'(x)@$ using chain rule.

By the way, there’s video version of this math tutorial:

As you know, first we need to define outside and inside functions, then find corresponding derivatives and multiply them. Start with @$f(x)@$:

$$f(x)=\sin^5{(3x+1) }$$

The inside function is @$3x+1@$. Then goes @$\sin{(3x+1)}@$ and finally the power @$\sin^5(3x+1)@$ . Applying chain rule for the case of three functions we get:

$$f'(x)=(\sin^5{(3x+1)})’=5\sin^4{(3x+1)} \cdot \cos{(3x+1)}\cdot 3=15\sin^4{(3x+1)}  \cos{(3x+1)}$$

That’s it. Now proceed to the second function:

$$g(x)=(4x+1)^3$$

The inside function here is @$4x+1@$, the outside function is power @$(4x+1)^3@$. Thus, we apply chain rule for the case of two functions and obtain:

$$g'(x)=((4x+1)^3)’=3(4x+1)^2\cdot 4=12(4x+1)^2$$

Now that we know all the required derivatives, we can substitute them into the formula for quotient rule:

$$\begin{split} h'(x)=(\frac{f(x)}{g(x)})’=\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2} =&\\\frac{15\sin^4{(3x+1)} \cos{(3x+1)} \cdot(4x+1)^3-\sin^5{(3x+1)}\cdot 12(4x+1)^2}{(4x+1)^3 )^2}=&\\\frac{3(4x+1)^2 \sin^4{(3x+1)}5\cos{(3x+1)}\cdot (4x+1)-4\sin{(3x+1)}}{(4x+1)^6}=&\\\frac{3\sin^4{(3x+1)} 5\cos{(3x+1)}\cdot (4x+1)-4 \sin{(3x+1)}}{(4x+1)^4}\end{split}$$

That’s our answer. If you face a function in your calculus homework which is a product or a quotient of other functions and components are also some compositions, remember that you need to use several rules for derivatives simultaneously.

Class vs homework

Also you can face a situation, where a function is a composition, and one of the components is a product. This may sound creepy, but basically you should act all the same except you need to go backwards, first applying product rule and then chain rule. Be attentive and you’ll find derivatives fast and easily.