Derive the function -1(x^2)/(y^2) with respect to x

\frac{d}{dx}\left(\left(-1\right)\cdot \frac{x^2}{y^2}\right)

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Answer

$-2xy^{-2}$

Step by step solution

Problem

$\frac{d}{dx}\left(\left(-1\right)\cdot \frac{x^2}{y^2}\right)$
1

The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function

$-\frac{d}{dx}\left(\frac{x^2}{y^2}\right)$
2

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$-\frac{y^2\frac{d}{dx}\left(x^2\right)-x^2\frac{d}{dx}\left(y^2\right)}{\left(y^2\right)^2}$
3

The derivative of the constant function is equal to zero

$-\frac{0\left(-1\right)x^2+y^2\frac{d}{dx}\left(x^2\right)}{\left(y^2\right)^2}$
4

Any expression multiplied by $0$ is equal to $0$

$-\frac{0+y^2\frac{d}{dx}\left(x^2\right)}{\left(y^2\right)^2}$
5

The power rule for differentiation states that if $n$ is a real number and $f(x) = x^n$, then $f'(x) = nx^{n-1}$

$-\frac{0+2xy^2}{\left(y^2\right)^2}$
6

$x+0=x$, where $x$ is any expression

$-\frac{2xy^2}{\left(y^2\right)^2}$
7

Applying the power of a power property

$-\frac{2xy^2}{y^{4}}$
8

Simplifying the fraction by $y$

$-1\cdot 2xy^{\left(2-4\right)}$
9

Subtract the values $2$ and $-4$

$-1\cdot 2xy^{-2}$
10

Multiply $2$ times $-1$

$-2xy^{-2}$
11

Applying the property of exponents, $\displaystyle a^{-n}=\frac{1}{a^n}$, where $n$ is a number

$-2x\frac{1}{y^{2}}$
12

Apply the formula: $a\frac{1}{x}$$=\frac{a}{x}$, where $a=-2$ and $x=y^{2}$

$x\frac{-2}{y^{2}}$
13

Multiplying the fraction and term

$\frac{-2x}{y^{2}}$
14

Rewrite the exponent using the power rule $\frac{a^m}{a^n}=a^{m-n}$, where in this case $m=0$

$-2xy^{-2}$

Answer

$-2xy^{-2}$

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Problem Analysis

Main topic:

Differential calculus

Time to solve it:

0.31 seconds

Views:

131