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# Find the implicit derivative $\frac{d}{dx}\left(\cos\left(xy\right)=1+\tan\left(y\right)\right)$

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##  Final answer to the problem

$y^{\prime}=\frac{-y\sin\left(xy\right)}{x\sin\left(xy\right)+\sec\left(y\right)^2}$
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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
Can't find a method? Tell us so we can add it.
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Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable

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

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

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

Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dx(cos(xy)=1+tan(y)). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. 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). Apply the product rule for differentiation: (f\cdot g)'=f'\cdot g+f\cdot g', where f=. The derivative of the linear function is equal to 1.

##  Final answer to the problem

$y^{\prime}=\frac{-y\sin\left(xy\right)}{x\sin\left(xy\right)+\sec\left(y\right)^2}$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

SnapXam A2

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0
a
b
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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

###  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.

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