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Find the implicit derivative $\frac{d}{dx}\left(\sqrt[3]{ax^2}+\sqrt[3]{by^2}=\sqrt[3]{cx}\right)$

Step-by-step Solution

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Final Answer

$\frac{2\sqrt[3]{a}}{3\sqrt[3]{x}}+\frac{2}{3\sqrt[3]{b^{2}}\sqrt[3]{y^{4}}}by\cdot y^{\prime}=\frac{\sqrt[3]{c}}{3\sqrt[3]{x^{2}}}$
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Step-by-step Solution

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

$\frac{d}{dx}\left(\sqrt[3]{ax^2}+\sqrt[3]{by^2}\right)=\frac{d}{dx}\left(\sqrt[3]{cx}\right)$

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

$\frac{d}{dx}\left(\sqrt[3]{ax^2}+\sqrt[3]{by^2}\right)=\frac{d}{dx}\left(\sqrt[3]{cx}\right)$

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Learn how to solve implicit differentiation problems step by step online. Find the implicit derivative d/dx((ax^2)^1/3+(by^2)^1/3=(cx)^1/3). Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable. The power rule for differentiation states that if n is a real number and f(x) = x^n, then f'(x) = nx^{n-1}. The derivative of a function multiplied by a constant (c) is equal to the constant times the derivative of the function. The derivative of the linear function is equal to 1.

Final Answer

$\frac{2\sqrt[3]{a}}{3\sqrt[3]{x}}+\frac{2}{3\sqrt[3]{b^{2}}\sqrt[3]{y^{4}}}by\cdot y^{\prime}=\frac{\sqrt[3]{c}}{3\sqrt[3]{x^{2}}}$

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Function Plot

Plotting: $\frac{2\sqrt[3]{a}}{3\sqrt[3]{x}}+\frac{2}{3\sqrt[3]{b^{2}}\sqrt[3]{y^{4}}}by\cdot y^{\prime}=\frac{\sqrt[3]{c}}{3\sqrt[3]{x^{2}}}$

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7
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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.

Used Formulas

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