# 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