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Find the implicit derivative $\frac{d}{dx}\left(\ln\left(xy\right)=e^{xy}\right)$

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Solution

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Final answer to the problem

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

$\frac{d}{dx}\left(\ln\left(xy\right)\right)=\frac{d}{dx}\left(e^{xy}\right)$
2

The derivative of the natural logarithm of a function is equal to the derivative of the function divided by that function. If $f(x)=ln\:a$ (where $a$ is a function of $x$), then $\displaystyle f'(x)=\frac{a'}{a}$

$\frac{1}{xy}\frac{d}{dx}\left(xy\right)=\frac{d}{dx}\left(e^{xy}\right)$
3

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

$\frac{1}{xy}\left(\frac{d}{dx}\left(x\right)y+x\frac{d}{dx}\left(y\right)\right)=\frac{d}{dx}\left(e^{xy}\right)$
4

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

$\frac{1}{xy}\left(y+xy^{\prime}\right)=\frac{d}{dx}\left(e^{xy}\right)$
5

Applying the derivative of the exponential function

$\frac{1}{xy}\left(y+xy^{\prime}\right)=e^{xy}\frac{d}{dx}\left(xy\right)$
6

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

$\frac{1}{xy}\left(y+xy^{\prime}\right)=e^{xy}\left(\frac{d}{dx}\left(x\right)y+x\frac{d}{dx}\left(y\right)\right)$
7

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

$\frac{1}{xy}\left(y+xy^{\prime}\right)=e^{xy}\left(y+xy^{\prime}\right)$
8

Multiply both sides of the equation by $xy$

$y+xy^{\prime}=e^{xy}\left(y+xy^{\prime}\right)xy$
9

Group the terms of the equation by moving the terms that have the variable $y^{\prime}$ to the left side, and those that do not have it to the right side

$xy^{\prime}-e^{xy}\left(y+xy^{\prime}\right)xy=-y$
10

Move everything to the left hand side of the equation

$xy^{\prime}-e^{xy}\left(y+xy^{\prime}\right)xy+y=0$
11

Factoring by $y+xy^{\prime}$

$\left(xy^{\prime}+y\right)\left(-e^{xy}xy+1\right)=0$
12

Break the equation in $2$ factors and set each factor equal to zero, to obtain simpler equations

$xy^{\prime}+y=0,\:-e^{xy}xy+1=0$
13

Solve the equation ($1$)

$xy^{\prime}+y=0$
14

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

$xy^{\prime}=-y$
15

Divide both sides of the equation by $x$

$y^{\prime}=\frac{-y}{x}$
16

Solve the equation ($2$)

$-e^{xy}xy+1=0$
17

This equation $-e^{xy}xy+1=0$ has no solutions in the real plane

$No solution$
18

The solution of the equation is

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

Final answer to the problem

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

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Function Plot

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

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0
a
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m
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u
v
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x
y
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.
(◻)
+
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×
◻/◻
/
÷
◻2
◻◻
√◻
√

◻√◻
◻√
∞
e
π
ln
log
log◻
lim
d/dx
D□x
∫
∫◻
|◻|
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

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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

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