# Derivative problems

### Derivative problems

Find the derivative of the following functions:

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

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

$$f(x) = \frac{x^3}{x + 1}, f'(x) = ?$$
Math Tutor

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### Re: Derivative problems

$$f(x) = 1, f'(x) = 0$$

$$f(x) = x^2 + 5x + 1, f'(x) = 2x+5$$

$$f(x) = \frac{x^3}{x + 1}, f'(x) = \frac{2x^3+3x^2}{(x+1)^2}$$

dduclam

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Excelent dduclam!

Math Tutor

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### Re: Derivative problems

The polynomial has two zeroes:$$x_1$$ and $$x_2$$.

$$x_2 = 7 * x_1$$

016hnoor

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### Re: Derivative problems

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

leesajohnson

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### Re: Derivative problems

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

There is a derivative tool that shows steps

Derivative solver

Math Tutor

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### Re: Derivative problems

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

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

### Re: Derivative problems

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[*]($$\frac{x^{2}+1}{x^{3}-x}$$)

the result should be f'(x)=$$\frac{e^{x}+3x^{2}}{e^{x}+x^{3}}$$)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

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### Re: Derivative problems

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 $$f(x)= log\left(\frac{x^2+1}{x^3- x}\right)$$) 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 $$ln(x)$$ is $$\frac{1}{x}$$. Next you need the chain rule: the derivative of $$f(u)$$ with u= g(x) is $$\frac{df}{du}\frac{du}{dx}$$. Putting $$u= \frac{x^2+1}{x^3- x}$$ then $$\frac{df}{dx}= \frac{1}{u}\frac{du}{dx}$$. And $$\frac{du}{dx}= \frac{d\left(\frac{x^2+1}{x^3-x}\right)}{dx}$$ is done using the quotient rule.

HallsofIvy

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