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

Step-by-step Solution

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Final answer to the problem

$y^{\prime}=\frac{-x}{y}$
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Step-by-step Solution

How should I solve this problem?

  • 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
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Can't find a method? Tell us so we can add it.
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

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

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

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

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

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

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

$2x+2y\cdot y^{\prime}=0$
7

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

$2y\cdot y^{\prime}=-2x$
8

Divide both sides of the equation by $2$

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

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

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

Divide both sides of the equation by $y$

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

Final answer to the problem

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

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Function Plot

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

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7
8
9
0
a
b
c
d
f
g
m
n
u
v
w
x
y
z
.
(◻)
+
-
×
◻/◻
/
÷
2

e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

How to improve your answer:

Main Topic: Implicit Differentiation

Implicit differentiation makes use of the chain rule to differentiate implicitly defined functions. For differentiating an implicit function y(x), defined by an equation R(x, y) = 0, it is not generally possible to solve it explicitly for y(x) and then differentiate. Instead, one can differentiate R(x, y) with respect to x and y and then solve a linear equation in dy/dx for getting explicitly the derivative in terms of x and y.

Used Formulas

See formulas (4)

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