# Find the derivative of x^3y^2*5+3y^5=pi+x3^0.5

## \frac{d}{dx}\left(5x^3 y^2+3y^5=\pi +x\sqrt{3}\right)

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$15y^2x^{2}=\sqrt{2}$

## Step by step solution

Problem

$\frac{d}{dx}\left(5x^3 y^2+3y^5=\pi +x\sqrt{3}\right)$
1

Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dx}\left(3y^5+5y^2x^3\right)=\frac{d}{dx}\left(\sqrt{2}x+\pi\right)$
2

The derivative of a sum of two functions is the sum of the derivatives of each function

$\frac{d}{dx}\left(3y^5\right)+\frac{d}{dx}\left(5y^2x^3\right)=\frac{d}{dx}\left(\sqrt{2}x\right)+\frac{d}{dx}\left(\pi\right)$
3

The derivative of the constant function is equal to zero

$0+\frac{d}{dx}\left(5y^2x^3\right)=\frac{d}{dx}\left(\sqrt{2}x\right)+0$
4

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

$0+\frac{d}{dx}\left(5y^2x^3\right)=\sqrt{2}\cdot\frac{d}{dx}\left(x\right)+0$
5

The derivative of the linear function is equal to $1$

$0+\frac{d}{dx}\left(5y^2x^3\right)=1\cdot \sqrt{2}+0$
6

Apply the product rule for differentiation: $(f\cdot g)'=f'\cdot g+f\cdot g'$, where $f=5x^3$ and $g=y^2$

$0+5x^3\frac{d}{dx}\left(y^2\right)+y^2\frac{d}{dx}\left(5x^3\right)=1\cdot \sqrt{2}+0$
7

The derivative of the constant function is equal to zero

$0+0\cdot 5x^3+y^2\frac{d}{dx}\left(5x^3\right)=1\cdot \sqrt{2}+0$
8

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

$0+0+y^2\frac{d}{dx}\left(5x^3\right)=1\cdot \sqrt{2}+0$
9

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

$0+0+5y^2\frac{d}{dx}\left(x^3\right)=1\cdot \sqrt{2}+0$
10

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

$0+0+5\cdot 3y^2x^{2}=1\cdot \sqrt{2}+0$
11

Add the values $0$ and $0$

$15y^2x^{2}=1\cdot \sqrt{2}+0$
12

Multiply $\sqrt{2}$ times $1$

$15y^2x^{2}=\sqrt{2}+0$
13

Add the values $0$ and $\sqrt{2}$

$15y^2x^{2}=\sqrt{2}$

$15y^2x^{2}=\sqrt{2}$

### Main topic:

Differential calculus

0.22 seconds

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