Find the derivative $\frac{d}{dx}\left(x+e^{\left(xy-2\right)}\right)$ using the sum rule

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Final answer to the problem

$1+e^{\left(xy-2\right)}y$
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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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The derivative of a sum of two or more functions is the sum of the derivatives of each function

$1+\frac{d}{dx}\left(e^{\left(xy-2\right)}\right)$

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$1+\frac{d}{dx}\left(e^{\left(xy-2\right)}\right)$

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Learn how to solve implicit differentiation problems step by step online. Find the derivative d/dx(x+e^(xy-2)) using the sum rule. The derivative of a sum of two or more functions is the sum of the derivatives of each function. Applying the derivative of the exponential function. The derivative of a sum of two or more functions is the sum of the derivatives of each function. The derivative of a function multiplied by a constant is equal to the constant times the derivative of the function.

Final answer to the problem

$1+e^{\left(xy-2\right)}y$

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Plotting: $1+e^{\left(xy-2\right)}y$

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Go!
1
2
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.

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