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

Step-by-step Solution

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Final Answer

$y^{\prime}=\frac{y\left(e^{\frac{x}{y}}x-1\right)}{x}$
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Step-by-step Solution

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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 $\displaystyle f'(x)=\frac{a'}{a}$

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

Learn how to solve implicit differentiation problems step by step online.

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

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Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dx(ln(xy))=e^(x/y). 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}. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=x and g=y. The derivative of the linear function is equal to 1. The derivative of the linear function is equal to 1.

Final Answer

$y^{\prime}=\frac{y\left(e^{\frac{x}{y}}x-1\right)}{x}$

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Function Plot

Plotting: $y^{\prime}=\frac{y\left(e^{\frac{x}{y}}x-1\right)}{x}$

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1
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3
4
5
6
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

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