Derivatives using product rule or qutoient rule

[conp]

I am struggling with understanding how to find derivatives using the product rule or quotient rule. Here are a few problems I am working on.

1.f(x)=x/e^x
2. z=1-t/1+t
3.w=3y+y^2/5+y
4. If f(x)=(3x+8)(2x-5) f ' (x) and f ' '(x)

5. if p is price in dollars and q is quantity, demand for a product is given by q=5000e^-0.08(p)
a)what quantity is sold at a price of $10?
b)find the derivative of demand with repect to the price when its 10.

 

[galactus]

I am struggling with understanding how to find derivatives using the product rule or quotient rule. Here are a few problems I am working on.

1.\(\displaystyle f(x)=\frac{x}{e^{x}}\)

You can use the product or quotient rule. Many problems can be converted from using the quotient rule to the product rule.

For instance, this one can be written as \(\displaystyle xe^{-x}\).

Let's step through both methods and see if we get the same result.

Product rule:

The first times the derivative of the second plus the second times the derivative of the first:

\(\displaystyle f(x)=\overbrace{x}^{\text{first}}\cdot \underbrace{e^{-x}}_{\text{second}}\)

\(\displaystyle \overbrace{x}^{\text{first}}\cdot \underbrace{(-e^{-x})}_{\text{times derivative of second}}+\overbrace{(e^{-x})}^{\text{second}}\cdot \underbrace{ (1)}_{\text{times derivative of first}}\)

Factor out \(\displaystyle e^{-x}\)

\(\displaystyle (1-x)e^{-x}\)

Quotient rule:

The denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator. All divided by the square of the denominator.

\(\displaystyle \frac{e^{x}(1)-x(e^{x})}{e^{2x}}=\frac{e^{x}(1-x)}{e^{2x}}=(1-x)e^{-x}\)

Same thing.

 

[soroban]

Hello, conp!

All of these involved the Quotient Rule. .Evidently, you're not familiar with it.
Galactus gave you an excellent explanation. .I hope you followed it.

I'll do #4 for you . . .

\(\displaystyle \text{4. If }f(x)\:=\:\frac{3x+8}{2x-5},\,\text{ find }f'(x)\text{ and }f''(x)\)

\(\displaystyle f'(x) \;=\;\frac{(2x-5)\cdot3 - (3x+8)\cdot2}{(2x-5)^2} \;=\;\frac{6x-15 - 6x - 16}{(2x-5)^2}\)

\(\displaystyle \text{Therefore: }\;\boxed{f'(x) \;=\;\frac{-31}{(2x-5)^2}}\)


\(\displaystyle \text{We have: }\:f'(x) \;=\;\text{-}31(2x-5)^{\text{-}2}\)

\(\displaystyle \text{Then: }\:f''(x) \;=\;\text{-}31(\text{-}2)(2x-5)^{\text{-}3}\cdot 2 \;=\; 124(2x-5)^{\text{-}3}\)

\(\displaystyle \text{Therefore: }\;\boxed{f''(x) \;=\;\frac{124}{(2x-5)^3}}\)

 

[conp]

Thanks for the explanation for both problems. I am now trying to attempt one of the problems by myself but I am not sure how to find the derivative of parts of the question.

ex. 1-t/1+t

I have (1+t)(1-t) - (1-t)(1+t)/(1+t)^2