Matias Aguirre

Answered

2022-07-19

How do you differentiate $x{e}^{{x}^{2}+{y}^{2}}$

Answer & Explanation

tun1t2j

Expert

2022-07-20Added 13 answers

To differentiate a two-variable function, you need to build the gradient. The gradient is a vector with as many coordinates as the variables the function depends on.

Each coordinate of the vector is a derivative with respect to one of the variables. So, in the two-variables case, you need to calculate the derivatives with respect to x and y, and then put them together in a vector.

Since deriving with respect to a variable means to consider the other as a constant, it's easier to derivate your function if it's expressed in the form

$x{e}^{{x}^{2}+{y}^{2}}=x\cdot {e}^{{x}^{2}}\cdot {e}^{{y}^{2}}$

So, deriving with respect to x, and using the product rule $\left(fg\right)\prime =f\prime g+fg\prime$ where $f\left(x\right)=x$ and $g\left(x\right)={e}^{{x}^{2}}$, we get

$\frac{d}{dx}x\cdot {e}^{{x}^{2}}\cdot {e}^{{y}^{2}}={e}^{{y}^{2}}({e}^{{x}^{2}}+x\cdot {e}^{{x}^{2}}\cdot 2x)=$

${e}^{{x}^{2}}\cdot {e}^{{y}^{2}}(1+2{x}^{2})={e}^{{x}^{2}+{y}^{2}}(1+2{x}^{2})$

Where the derivative of $e}^{{x}^{2}$ has been calculated using the chain rule, which states that $f\left(g\left(x\right)\right)\prime =f\prime \left(g\left(x\right)\right)\cdot g\prime \left(x\right)$, where $f\left(x\right)={e}^{x}$, and $g\left(x\right)={x}^{2}$

The derivative with respect to y is easier, since the only factor to differentiate is $e}^{{y}^{2}$, while the others depend only on x and are thus to be consideblack as constant. So, we have

$\frac{d}{dy}x\cdot {e}^{{x}^{2}}\cdot {e}^{{y}^{2}}=x\cdot {e}^{{x}^{2}}\cdot {e}^{{y}^{2}}\cdot 2y=$

$2xy{e}^{{x}^{2}+{y}^{2}}$

The gradient is thus the vector

$({e}^{{x}^{2}+{y}^{2}}(1+2{x}^{2}),2xy{e}^{{x}^{2}+{y}^{2}})$

Each coordinate of the vector is a derivative with respect to one of the variables. So, in the two-variables case, you need to calculate the derivatives with respect to x and y, and then put them together in a vector.

Since deriving with respect to a variable means to consider the other as a constant, it's easier to derivate your function if it's expressed in the form

$x{e}^{{x}^{2}+{y}^{2}}=x\cdot {e}^{{x}^{2}}\cdot {e}^{{y}^{2}}$

So, deriving with respect to x, and using the product rule $\left(fg\right)\prime =f\prime g+fg\prime$ where $f\left(x\right)=x$ and $g\left(x\right)={e}^{{x}^{2}}$, we get

$\frac{d}{dx}x\cdot {e}^{{x}^{2}}\cdot {e}^{{y}^{2}}={e}^{{y}^{2}}({e}^{{x}^{2}}+x\cdot {e}^{{x}^{2}}\cdot 2x)=$

${e}^{{x}^{2}}\cdot {e}^{{y}^{2}}(1+2{x}^{2})={e}^{{x}^{2}+{y}^{2}}(1+2{x}^{2})$

Where the derivative of $e}^{{x}^{2}$ has been calculated using the chain rule, which states that $f\left(g\left(x\right)\right)\prime =f\prime \left(g\left(x\right)\right)\cdot g\prime \left(x\right)$, where $f\left(x\right)={e}^{x}$, and $g\left(x\right)={x}^{2}$

The derivative with respect to y is easier, since the only factor to differentiate is $e}^{{y}^{2}$, while the others depend only on x and are thus to be consideblack as constant. So, we have

$\frac{d}{dy}x\cdot {e}^{{x}^{2}}\cdot {e}^{{y}^{2}}=x\cdot {e}^{{x}^{2}}\cdot {e}^{{y}^{2}}\cdot 2y=$

$2xy{e}^{{x}^{2}+{y}^{2}}$

The gradient is thus the vector

$({e}^{{x}^{2}+{y}^{2}}(1+2{x}^{2}),2xy{e}^{{x}^{2}+{y}^{2}})$

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