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Exercise

$\frac{d}{dx}\left(x^2+y^2=25\right)$

Step-by-step Solution

1

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

$\frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(25\right)$
2

The derivative of the constant function ($25$) is equal to zero

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

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

Make diagrams like this with our diagram generator
$\frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)=0$
4

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

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

Any expression to the power of $1$ is equal to that same expression

$\frac{d}{dx}\left(x^2\right)+2y\frac{d}{dx}\left(y\right)=0$
6

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

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

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

Make diagrams like this with our diagram generator
$2x+2y\cdot y^{\prime}=0$
8

We need to isolate the dependent variable $y$, we can do that by simultaneously subtracting $2x$ from both sides of the equation

$2y\cdot y^{\prime}=-2x$
9

Divide both sides of the equation by $2$

$y^{\prime}y=\frac{-2x}{2}$
10

Take $\frac{-2}{2}$ out of the fraction

$y^{\prime}y=-x$
11

Divide both sides of the equation by $y$

$y^{\prime}=\frac{-x}{y}$

Final answer to the exercise

$y^{\prime}=\frac{-x}{y}$

Try other ways to solve this exercise

  • Choose an option
  • Find the derivative using the definition
  • Find the derivative using the product rule
  • Find the derivative using the quotient rule
  • Find the derivative using logarithmic differentiation
  • Find the derivative
  • Integrate by partial fractions
  • Product of Binomials with Common Term
  • FOIL Method
  • Integrate by substitution
  • Load more...
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