Derivative problems

[Math Tutor]

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]

[dduclam]

[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]

[Math Tutor]

Excelent dduclam!

[016hnoor]

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

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

[leesajohnson]

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

[Guest]

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.

[Math Tutor]

There is a derivative tool that shows steps

Derivative solver

[Guest]

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].

[Drey999]

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.

[HallsofIvy]

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.

[Guest]

Hi, how can one differentiate both sides of this equation with respect to z? that's what I have been told to do but I am not sure how to do that.
[tex]\frac{dg(z,t)}{dt}=Qz(z-1)\frac{dg(z,t)}{dz}[/tex]

[HallsofIvy]

Hi, how can one differentiate both sides of this equation with respect to z? that's what I have been told to do but I am not sure how to do that.
[tex]\frac{dg(z,t)}{dt}=Qz(z-1)\frac{dg(z,t)}{dz}[/tex]

First, because this involves a function of two variables, z and t, and has derivatives with respect to each variable, I am going to write it with partial derivative
[tex]\frac{\partial g(z,t)}{\partial t}= Qz(z- 1)\frac{\partial g(z,t)}{\partial z}[/tex]

The derivative on the left is just [tex]\frac{\partial^2g}{\partial t\partial z}[/tex]

On the right, assuming Q is a constant we have to use the "product rule" with z, z- 1, and [tex]\frac{\partial g(z,t)}{\partial z}[/tex] so the derivative with respect to z is
[tex]Q(z- 1)\frac{\partial g(z,t)}{\partial z}+ Qz\frac{\partial g(z,t)}{\partial z}+ Qz(z- 1)\frac{\partial^2 g(z,t)}{\partial z^2}[/tex].

Putting those together
[tex]\frac{\partial^2g}{\partial t\partial z}= Q(z- 1)\frac{\partial g(z,t)}{\partial z}+ Qz\frac{\partial g(z,t)}{\partial z}+ Qz(z- 1)\frac{\partial^2 g(z,t)}{\partial z^2}[/tex].

[HallsofIvy]

Hi, how can one differentiate both sides of this equation with respect to z? that's what I have been told to do but I am not sure how to do that.
[tex]\frac{dg(z,t)}{dt}=Qz(z-1)\frac{dg(z,t)}{dz}[/tex]

First, because this involves a function of two variables, z and t, and has derivatives with respect to each variable, I am going to write it with partial derivatives
[tex]\frac{\partial g(z,t)}{\partial t}= Qz(z- 1)\frac{\partial g(z,t)}{\partial z}[/tex].
(The Latex code for [tex]\partial[/tex] is "\partial".)

The derivative on the left is just [tex]\frac{\partial^2g(z,t)}{\partial t\partial z}[/tex].

On the right, assuming Q is a constant we have to use the "product rule" with z, z- 1, and [tex]\frac{\partial g(z,t)}{\partial z}[/tex] so the derivative with respect to z is
[tex]Q(z- 1)\frac{\partial g(z,t)}{\partial z}+ Qz\frac{\partial g(z,t)}{\partial z}+ Qz(z- 1)\frac{\partial^2 g(z,t)}{\partial z^2}[/tex].

Putting those together
[tex]\frac{\partial^2g}{\partial t\partial z}= Q(z- 1)\frac{\partial g(z,t)}{\partial z}+ Qz\frac{\partial g(z,t)}{\partial z}+ Qz(z- 1)\frac{\partial^2 g(z,t)}{\partial z^2}[/tex].