Inverse trigonometric functions differentiation Calculator
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$\frac{d}{dx}\left(arctan\left(\frac{x}{y}\right)\right)$
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Taking the derivative of arctangent
$\frac{1}{\left(\frac{x}{y}\right)^2+1}\cdot\frac{d}{dx}\left(\frac{x}{y}\right)$
3
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 ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$
$\frac{1}{\left(\frac{x}{y}\right)^2+1}\cdot\frac{y\frac{d}{dx}\left(x\right)-x\frac{d}{dx}\left(y\right)}{y^2}$
4
The derivative of the constant function is equal to zero
$\frac{1}{\left(\frac{x}{y}\right)^2+1}\cdot\frac{y\frac{d}{dx}\left(x\right)-0x}{y^2}$
5
Any expression multiplied by $0$ is equal to $0$
$0$
6
The derivative of the linear function is equal to $1$
$\frac{0+y}{y^2}\cdot\frac{1}{\left(\frac{x}{y}\right)^2+1}$
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$x+0=x$, where $x$ is any expression
$\frac{y}{y^2}\cdot\frac{1}{\left(\frac{x}{y}\right)^2+1}$
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Simplifying the fraction by $y$
$\frac{1}{y}\cdot\frac{1}{\left(\frac{x}{y}\right)^2+1}$
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The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$
$\frac{1}{y}\cdot\frac{1}{\frac{x^2}{y^2}+1}$
$\frac{1}{y\left(\frac{x^2}{y^2}+1\right)}$
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Multiplying polynomials $y$ and $1+\frac{x^2}{y^2}$
$\frac{1}{\frac{x^2y}{y^2}+y}$
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Simplifying the fraction by $y$
$\frac{1}{\frac{x^2}{y}+y}$
13
Apply the formula: $\frac{b}{c}+a$$=\frac{b+c\cdot a}{c}$, where $a=y$, $b=x^2$ and $c=y$
$\frac{1}{\frac{x^2+y\cdot y}{y}}$
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When multiplying exponents with same base you can add the exponents
$\frac{1}{\frac{x^2+y^2}{y}}$
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Simplifying the fraction
$\frac{y}{x^2+y^2}$