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Here, we show you a step-by-step solved example of implicit differentiation. This solution was automatically generated by our smart calculator:
d d x ( x 2 + y 2 = 16 ) \frac{d}{dx}\left(x^2+y^2=16\right) d x d β ( x 2 + y 2 = 16 )
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Apply implicit differentiation by taking the derivative of both sides of the equation with respect to the differentiation variable
d d x ( x 2 + y 2 ) = d d x ( 16 ) \frac{d}{dx}\left(x^2+y^2\right)=\frac{d}{dx}\left(16\right) d x d β ( x 2 + y 2 ) = d x d β ( 16 )
3
The derivative of the constant function (16 16 16
d d x ( x 2 + y 2 ) = 0 \frac{d}{dx}\left(x^2+y^2\right)=0 d x d β ( x 2 + y 2 ) = 0
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The derivative of a sum of two or more functions is the sum of the derivatives of each function
d d x ( x 2 ) + d d x ( y 2 ) = 0 \frac{d}{dx}\left(x^2\right)+\frac{d}{dx}\left(y^2\right)=0 d x d β ( x 2 ) + d x d β ( y 2 ) = 0
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Intermediate steps
The power rule for differentiation states that if n n n f ( x ) = x n f(x) = x^n f ( x ) = x n f β² ( x ) = n x n β 1 f'(x) = nx^{n-1} f β² ( x ) = n x n β 1
d d x ( x 2 ) + 2 y 2 β 1 d d x ( y ) = 0 \frac{d}{dx}\left(x^2\right)+2y^{2-1}\frac{d}{dx}\left(y\right)=0 d x d β ( x 2 ) + 2 y 2 β 1 d x d β ( y ) = 0
Add the values 2 2 2 β 1 -1 β 1
d d x ( x 2 ) + 2 y 1 d d x ( y ) = 0 \frac{d}{dx}\left(x^2\right)+2y^{1}\frac{d}{dx}\left(y\right)=0 d x d β ( x 2 ) + 2 y 1 d x d β ( y ) = 0
The power rule for differentiation states that if n n n f ( x ) = x n f(x) = x^n f ( x ) = x n f β² ( x ) = n x n β 1 f'(x) = nx^{n-1} f β² ( x ) = n x n β 1
2 y 2 β 1 d d x ( y ) 2y^{2-1}\frac{d}{dx}\left(y\right) 2 y 2 β 1 d x d β ( y )
Subtract the values 2 2 2 β 1 -1 β 1
2 y 1 d d x ( y ) 2y^{1}\frac{d}{dx}\left(y\right) 2 y 1 d x d β ( y )
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The power rule for differentiation states that if n n n f ( x ) = x n f(x) = x^n f ( x ) = x n f β² ( x ) = n x n β 1 f'(x) = nx^{n-1} f β² ( x ) = n x n β 1
d d x ( x 2 ) + 2 y 1 d d x ( y ) = 0 \frac{d}{dx}\left(x^2\right)+2y^{1}\frac{d}{dx}\left(y\right)=0 d x d β ( x 2 ) + 2 y 1 d x d β ( y ) = 0
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Explain this step further
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Any expression to the power of 1 1 1
d d x ( x 2 ) + 2 y d d x ( y ) = 0 \frac{d}{dx}\left(x^2\right)+2y\frac{d}{dx}\left(y\right)=0 d x d β ( x 2 ) + 2 y d x d β ( y ) = 0
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The derivative of the linear function is equal to 1 1 1
d d x ( x 2 ) + 2 y β
y β² = 0 \frac{d}{dx}\left(x^2\right)+2y\cdot y^{\prime}=0 d x d β ( x 2 ) + 2 y β
y β² = 0
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Intermediate steps
The power rule for differentiation states that if n n n f ( x ) = x n f(x) = x^n f ( x ) = x n f β² ( x ) = n x n β 1 f'(x) = nx^{n-1} f β² ( x ) = n x n β 1
2 x ( 2 β 1 ) 2x^{\left(2-1\right)} 2 x ( 2 β 1 )
Subtract the values 2 2 2 β 1 -1 β 1
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The power rule for differentiation states that if n n n f ( x ) = x n f(x) = x^n f ( x ) = x n f β² ( x ) = n x n β 1 f'(x) = nx^{n-1} f β² ( x ) = n x n β 1
2 x + 2 y β
y β² = 0 2x+2y\cdot y^{\prime}=0 2 x + 2 y β
y β² = 0
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Explain this step further
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We need to isolate the dependent variable y y y 2 x 2x 2 x
2 y β
y β² = β 2 x 2y\cdot y^{\prime}=-2x 2 y β
y β² = β 2 x
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Divide both sides of the equation by 2 2 2
y β² y = β 2 x 2 y^{\prime}y=\frac{-2x}{2} y β² y = 2 β 2 x β
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Take β 2 2 \frac{-2}{2} 2 β 2 β
y β² y = β x y^{\prime}y=-x y β² y = β x
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Divide both sides of the equation by y y y
y β² = β x y y^{\prime}=\frac{-x}{y} y β² = y β x β
ξ Final answer to the exercise
y β² = β x y y^{\prime}=\frac{-x}{y} y β² = y β x β ξ