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

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Final answer to the problem

$y^{\prime}+1=\left(y^2+2xy\cdot y^{\prime}\right)\cos\left(xy^2\right)$
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Step-by-step Solution

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  • 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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Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

$\frac{d}{dy}\left(y+x\right)=\frac{d}{dy}\left(\sin\left(xy^2\right)\right)$

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

$\frac{d}{dy}\left(y+x\right)=\frac{d}{dy}\left(\sin\left(xy^2\right)\right)$

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Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dy(y+x=sin(xy^2)). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. The derivative of the sine of a function is equal to the cosine of that function times the derivative of that function, in other words, if {f(x) = \sin(x)}, then {f'(x) = \cos(x)\cdot D_x(x)}. Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=x and g=y^2. The derivative of the linear function is equal to 1.

Final answer to the problem

$y^{\prime}+1=\left(y^2+2xy\cdot y^{\prime}\right)\cos\left(xy^2\right)$

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Function Plot

Plotting: $y^{\prime}+1=\left(y^2+2xy\cdot y^{\prime}\right)\cos\left(xy^2\right)$

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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
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 (5)

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