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3.5 Selecting Procedures for Calculating Derivatives

1 min readjune 18, 2024

3.5 Selecting Procedures for Calculating Derivatives

Congratulations! 👏 You are now a pro at calculating derivatives! Over the past two units, you have learned many derivative rules, from the power rule to the chain rule. This knowledge has cumulated in your ability to perform advanced derivative problems involving composite, implicit, and inverse functions. So pat yourself on the back! 🤗

📝 Calculating Derivatives Practice

To effectively solve derivative problems, it‘s important to first recognize which procedures to use. Below are some practice problems to help us do such!

❓ Calculating Derivatives Practice Problems

Question 1:

Which sequence of rules can be used to differentiate f(x)=cos(x3)5xf(x) = \frac{\cos(x^{3})}{5x}?

A) Quotient rule, then quotient rule again

B) Quotient rule, then chain rule

C) Chain rule, then chain rule again

D) Quotient rule, then product rule

Question 2:

Which sequence of rules can be used to differentiate g(x)=4xcos(x)sin(x)g(x) = 4x\cos(x)\sin(x)?

A) Chain rule, then product rule

B) Chain rule, then chain rule again

C) Product rule, then chain rule

D) Product rule, then product rule again

Question 3:

What is the derivative of h(x)=(3x315x)(2xx7)h(x) = (3x^{3}-15x)(2x-x^{7})?

Question 4:

What is the derivative of f(x)=6ex3+4f(x)= 6e^{x^3+4}?


✅ Answers and Solutions

Question 1: Differentiating f(x)=cos(x3)5xf(x) = \frac{\cos(x^{3})}{5x}

Answer: B) Quotient rule, then chain rule

Explanation:

f(x)=cos(x3)5xf(x) = \frac{\cos(x^{3})}{5x} is the quotient of one function, cos(x3)\cos(x^{3}), and another function, 5x5x. This indicates that the Quotient Rule should be used.

The function cos(x3)\cos(x^{3}) is a composite function with the outside function being cos(x)\cos(x) and the inside function being x3x^{3}. This means that to find its derivative, the Chain Rule needs to be applied.

The sequence of rules that should be used to differentiate f(x)f(x) is B) Quotient rule, then the chain rule.


Question 2: Differentiating g(x)=4xcos(x)sin(x)g(x) = 4x\cos(x)\sin(x)

Answer: D) Product rule, then product rule again

Explanation:

g(x)=4xcos(x)sin(x)g(x) = 4x\cos(x)\sin(x) is the product of three functions, 4x4x, cos(x)\cos(x), and sin(x)\sin(x). Therefore, the Product Rule needs to be applied twice to find its derivative, making D the correct answer.


Question 3: Differentiating h(x)=(3x315x)(2xx7)h(x) = (3x^{3}-15x)(2x-x^{7})

Answer: 30x9+120x7+24x360x-30x^{9}+120x^{7}+24x^{3}-60x

Solution:

h(x)=(3x315x)(2xx7)h(x) = (3x^{3}-15x)(2x-x^{7}) is the product of two functions, 3x315x3x^{3}-15x and 2xx72x-x^{7}. This means that we need to apply the Product Rule:

ddx[f(x)g(x)]=f(x)g(x)+f(x)g(x)\frac{d}{dx}[f(x)g(x)]=f'(x)g(x)+f(x)g'(x)

The derivative of 3x315x3x^{3}-15x is 9x2159x^{2}-15, by the Power Rule and the Constant Multiple Rule.

The derivative of 2xx72x-x^{7} is 27x62-7x^{6}, by the Power Rule and the Constant Multiple Rule.

So, by the Product Rule, h(x)=(9x215)(2xx7)+(3x315x)(27x6)h'(x) = (9x^{2}-15)(2x-x^{7})+(3x^{3}-15x)(2-7x^{6}).

This simplifies to 30x9+120x7+24x360x-30x^{9}+120x^{7}+24x^{3}-60x, which is our answer.


Question 4: Differentiating f(x)=6ex3+4f(x)= 6e^{x^3+4}

Answer: 18ex3+4x218e^{x^3+4}x^2

Solution:

f(x)=6ex3+4f(x)= 6e^{x^3+4} is a composite function with the outside function being exe^{x} and the inside function being x3+4x^3+4 (we can ignore the 66 for now because we can simply multiply it at the end based on the Constant Multiple Rule).

This means that to find its derivative, we need to apply the Chain Rule:

ddx[f(g(x))]=f(g(x))g(x)\frac{d}{dx}[f(g(x))]=f'(g(x)) \cdot g'(x)

The derivative of exe^{x} is exe^{x}, so f(g(x))f'(g(x)) is ex3+4e^{x^3+4}.

The derivative of x3+4x^3+4 is 3x23x^2.

So, by the Chain Rule, f(x)=6ex3+43x2f'(x)=6e^{x^3+4}\cdot 3x^2, which simplifies to 18ex3+4x218e^{x^3+4}x^2.


These are just a few questions to get you thinking about what rules you should be using when faced with calculating derivatives! The more you practice, the easier it’ll become. You got this! 🍀