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Find the implicit derivative $\frac{d}{dx}\left(x+\cos\left(x+y\right)\right)=0$

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Solution

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Final answer to the problem

$1-\left(1+y^{\prime}\right)\sin\left(x+y\right)=0$
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Step-by-step Solution

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  • Find the derivative using the definition
  • Exact Differential Equation
  • Linear Differential Equation
  • Separable Differential Equation
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  • Find the derivative using the quotient rule
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1

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

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

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

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Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dx(x+cos(x+y))=0. The derivative of a sum of two or more functions is the sum of the derivatives of each function. The derivative of the cosine of a function is equal to minus the sine of the function times the derivative of the function, in other words, if f(x) = \cos(x), then f'(x) = -\sin(x)\cdot D_x(x). The derivative of a sum of two or more functions is the sum of the derivatives of each function.

Final answer to the problem

$1-\left(1+y^{\prime}\right)\sin\left(x+y\right)=0$

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Plotting: $1-\left(1+y^{\prime}\right)\sin\left(x+y\right)=0$

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0
a
b
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d
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m
n
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v
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x
y
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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

See formulas (3)

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