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Find the implicit derivative $\frac{d}{dx}\left(\ln\left(x+y\right)\right)=4$

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Solution

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Final answer to the problem

$y^{\prime}=4x+4y-1$
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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 $\displaystyle f'(x)=\frac{a'}{a}$

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

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

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

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Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dx(ln(x+y))=4. 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}. The derivative of a sum of two or more functions is the sum of the derivatives of each function. Multiply both sides of the equation by x+y. Any expression multiplied by 1 is equal to itself.

Final answer to the problem

$y^{\prime}=4x+4y-1$

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Function Plot

Plotting: $y^{\prime}=4x+4y-1$

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