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]
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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:

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

Excelent dduclam!

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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]

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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

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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.
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Re: Derivative problems

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

There is a derivative tool that shows steps

Derivative solver

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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.
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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].
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