Derivative problems

Derivative problems

Postby Math Tutor » Tue Jan 08, 2008 11:41 am

Find the derivative of the following functions:

f(x) = 1, f'(x) = ?

f(x) = x2 + 5x + 1, f'(x) = ?

[tex]f(x) = \frac{x^3}{x + 1}, f'(x) = ?[/tex]
Math Tutor
Site Admin
 
Posts: 410
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 28

Re: Derivative problems

Postby dduclam » Tue Mar 25, 2008 4:53 am



[tex]f(x) = 1, f'(x) = 0[/tex]

[tex]f(x) = x^2 + 5x + 1, f'(x) = 2x+5[/tex]

[tex]f(x) = \frac{x^3}{x + 1}, f'(x) = \frac{2x^3+3x^2}{(x+1)^2}[/tex]

:wink:

dduclam
 
Posts: 36
Joined: Sat Dec 29, 2007 10:42 am
Location: HUCE-Vietnam
Reputation: 3

Postby Math Tutor » Tue Mar 25, 2008 8:50 am

Excelent dduclam!

Math Tutor
Site Admin
 
Posts: 410
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 28

Re: Derivative problems

Postby 016hnoor » Thu Mar 19, 2015 1:04 am

The polynomial has two zeroes:[tex]x_1[/tex] and [tex]x_2[/tex].

[tex]x_2 = 7 * x_1[/tex]

016hnoor
 
Posts: 3
Joined: Thu Mar 19, 2015 12:41 am
Reputation: 1

Re: Derivative problems

Postby leesajohnson » Wed Jan 27, 2016 8:15 am

The derivative of the function is
f(x)=x³/x+1, f′(x)= 2x³+3x²/(x+1)²

User avatar
leesajohnson
 
Posts: 208
Joined: Thu Dec 31, 2015 7:11 am
Location: London
Reputation: -33

Re: Derivative problems

Postby Guest » Tue Oct 30, 2018 8:21 am

Guys, huge problem: what is the derivative from this
3x^2*y^2-5x+siny=(3y-1)^4


I really need to solve this, but i don't even know where to start.
Guest
 

Re: Derivative problems

Postby Math Tutor » Tue Oct 30, 2018 8:27 am

There is a derivative tool that shows steps

Derivative solver

Math Tutor
Site Admin
 
Posts: 410
Joined: Sun Oct 09, 2005 11:37 am
Reputation: 28

Re: Derivative problems

Postby Guest » Wed Aug 21, 2019 7:37 am

Guys, huge problem: what is the derivative from this
3x^2*y^2-5x+siny=(3y-1)^4


I really need to solve this, but i don't even know where to start.
Guest


First of all, that makes no sense. You find the derivative of a function, not an equation. I will assume that you are to find the derivative of y with respect to x, dy/dx.

Using the chain rule, the derivative of [tex]y^2[/tex] with respect to x is the derivative of [tex]y^2[/tex] with respect to y times the derivative of y with respect to x: [tex]2y\frac{dy}{dx}[/tex] so the derivative of [tex]3x^2y^2[/tex] with respect to x is, by the product rule, [tex]6xy^2+ 6x^2y\frac{dy}{dx}[/tex]. Similarly the derivative of [tex]sin(y)[/tex] with respect to x is [tex]cos(y)\frac{dy}{dx}[/tex] and the derivative of [tex](3y- 1)^4[/tex] with respect to x is [tex]4(3y- 1)^3(3)\frac{dy}{dx}= 12(3y- 1)^3\frac{dy}{dx}[/tex]

So, differentiating both sides of [tex]3x^2*y^2-5x+siny=(3y-1)^4[/tex] with respect to x we get [tex]6xy^2+ 6x^2y\frac{dy}{dx}- 5+ cos(y)\frac{dy}{dx}= 12(3y- 1)^3\frac{dy}{dx}[/tex]. Finally, solve that for [tex]\frac{dy}{dx}[/tex].
Guest
 

Re: Derivative problems

Postby Drey999 » Mon Oct 21, 2019 1:34 pm

Greetings.
I have been practicing differentiation logarithmic and exponential functions when i encountered an exercise that said to find the derivative of:
f(x)=ln[*]([tex]\frac{x^{2}+1}{x^{3}-x}[/tex])

the result should be f'(x)=[tex]\frac{e^{x}+3x^{2}}{e^{x}+x^{3}}[/tex])according to the answer sheet but i am confused on how to get there, I've tried to start by simplifying the function followed by the chain rule but it didn't give me the ordered result can someone please help me solving this question.
thank you and have a great day.

Drey999
 
Posts: 1
Joined: Mon Oct 21, 2019 1:18 pm
Reputation: 0

Re: Derivative problems

Postby HallsofIvy » Mon Dec 02, 2019 2:29 pm

I suspect you have either stated the problem incorrectly (I don't know what "log[*]" is supposed to mean. I am going to assume that [tex]f(x)= log\left(\frac{x^2+1}{x^3- x}\right)[/tex]) or looked up the wrong answer! The function you say is the derivative can't possibly be correct because you can't get exponentials as derivatives of anything except exponentials!

For this problem you need to know that the derivative of [tex]ln(x)[/tex] is [tex]\frac{1}{x}[/tex]. Next you need the chain rule: the derivative of [tex]f(u)[/tex] with u= g(x) is [tex]\frac{df}{du}\frac{du}{dx}[/tex]. Putting [tex]u= \frac{x^2+1}{x^3- x}[/tex] then [tex]\frac{df}{dx}= \frac{1}{u}\frac{du}{dx}[/tex]. And [tex]\frac{du}{dx}= \frac{d\left(\frac{x^2+1}{x^3-x}\right)}{dx}[/tex] is done using the quotient rule.

HallsofIvy
 
Posts: 81
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 27


Return to Functions, Graphs, Derivatives



Who is online

Users browsing this forum: No registered users and 1 guest

cron