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# find the derivative using implicit differentiation. tan(x − y) = y/(5 + x^2)

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find the derivative using implicit differentiation. tan(x − y) = y/(5 + x^2)

-please add explanation. i get stuck when trying to bring dy/dx to one side of the equation.

Feb 26, 2020

#1
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Find the derivative of the following via implicit differentiation:
d/dx(tan(x - y)) = d/dx(y/(5 + x^2))

Using the chain rule, d/dx(tan(x - y)) = ( dtan(u))/( du) ( du)/( dx), where u = x - y and d/( du)(tan(u)) = sec^2(u):
(d/dx(x - y)) sec^2(x - y) = d/dx(y/(5 + x^2))

Differentiate the sum term by term and factor out constants:
d/dx(x) - d/dx(y) sec^2(x - y) = d/dx(y/(5 + x^2))

The derivative of x is 1:
sec^2(x - y) (-(d/dx(y)) + 1) = d/dx(y/(5 + x^2))

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):
sec^2(x - y) (1 - (d/dx(x)) y'(x)) = d/dx(y/(5 + x^2))

The derivative of x is 1:
sec^2(x - y) (1 - 1 y'(x)) = d/dx(y/(5 + x^2))

Use the quotient rule, d/dx(u/v) = (v ( du)/( dx) - u ( dv)/( dx))/v^2, where u = y and v = x^2 + 5:
sec^2(x - y) (1 - y'(x)) = ((5 + x^2) (d/dx(y)) - y (d/dx(5 + x^2)))/(5 + x^2)^2

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):
sec^2(x - y) (1 - y'(x)) = ((5 + x^2) (d/dx(x)) y'(x) - (d/dx(5 + x^2)) y)/(5 + x^2)^2

The derivative of x is 1:
sec^2(x - y) (1 - y'(x)) = (-(d/dx(5 + x^2)) y + 1 (5 + x^2) y'(x))/(5 + x^2)^2

Differentiate the sum term by term:
sec^2(x - y) (1 - y'(x)) = (-y d/dx(5) + d/dx(x^2) + (5 + x^2) y'(x))/(5 + x^2)^2

The derivative of 5 is zero:
sec^2(x - y) (1 - y'(x)) = (-(d/dx(x^2) + 0) y + (5 + x^2) y'(x))/(5 + x^2)^2

Simplify the expression:
sec^2(x - y) (1 - y'(x)) = (-(d/dx(x^2)) y + (5 + x^2) y'(x))/(5 + x^2)^2

Use the power rule, d/dx(x^n) = n x^(n - 1), where n = 2.
d/dx(x^2) = 2 x:
sec^2(x - y) (1 - y'(x)) = (-y 2 x + (5 + x^2) y'(x))/(5 + x^2)^2

Expand the left hand side:
sec^2(x - y) - sec^2(x - y) y'(x) = (-2 x y + (5 + x^2) y'(x))/(5 + x^2)^2

Expand the right hand side:
sec^2(x - y) - sec^2(x - y) y'(x) = -(2 x y)/(5 + x^2)^2 + (5 y'(x))/(5 + x^2)^2 + (x^2 y'(x))/(5 + x^2)^2

Subtract (x^2 y'(x))/(x^2 + 5)^2 + (5 y'(x))/(x^2 + 5)^2 from both sides:
sec^2(x - y) - (5 y'(x))/(5 + x^2)^2 - (x^2 y'(x))/(5 + x^2)^2 - sec^2(x - y) y'(x) = -(2 x y)/(5 + x^2)^2

Subtract sec^2(x - y) from both sides:
-(5 y'(x))/(5 + x^2)^2 - (x^2 y'(x))/(5 + x^2)^2 - sec^2(x - y) y'(x) = -sec^2(x - y) - (2 x y)/(5 + x^2)^2

Collect the left hand side in terms of y'(x):
(-5/(5 + x^2)^2 - x^2/(5 + x^2)^2 - sec^2(x - y)) y'(x) = -sec^2(x - y) - (2 x y)/(5 + x^2)^2

Divide both sides by -x^2/(x^2 + 5)^2 - 5/(x^2 + 5)^2 - sec^2(x - y):

y'(x) = (-sec^2(x - y) - (2 x y)/(5 + x^2)^2)/(-5/(5 + x^2)^2 - x^2/(5 + x^2)^2 - sec^2(x - y))

Courtesy of Mathematica 11 Home Edition.

Feb 26, 2020
#2
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find the derivative using implicit differentiation. $$\tan(x - y) = \dfrac{y}{5 + x^2}$$

Derivative: $$(5 + x^2)\tan(x - y) -y=0$$

Formula: $$y'(x) = -\dfrac{F_x}{F_y}$$

$$\small{ \begin{array}{|rcll|} \hline \mathbf{F_x = \ ?} \\ \hline F_x &=& \dfrac{d }{dx}(5 + x^2)\tan(x - y)- \dfrac{d}{dx}y \quad | \quad \dfrac{d}{dx}y = 0 \\ F_x &=& \dfrac{d }{dx}(5 + x^2)\tan(x - y)-0 \\ F_x &=& \dfrac{d }{dx}(5 + x^2)\tan(x - y) \\ F_x &=& (5 + x^2)\dfrac{d }{dx}\tan(x - y)+\tan(x - y)\dfrac{d }{dx}(5 + x^2) \quad | \quad \dfrac{d }{dx}(5 + x^2) = 2x \\ F_x &=& (5 + x^2)\dfrac{d }{dx}\tan(x - y)+2x\tan(x - y)\dfrac{d }{dx} \quad | \quad \dfrac{d }{dx}\tan(x - y) = \dfrac{1}{\cos^2(x-y)}*1 \\ F_x &=& (5 + x^2)\dfrac{1}{\cos^2(x-y)}*1+2x\tan(x - y) \\ F_x &=& \dfrac{5 + x^2+2x\tan(x - y)\cos^2(x-y)}{\cos^2(x-y)} \\ F_x &=& \dfrac{5 + x^2+2x\sin(x - y)\cos(x-y)}{\cos^2(x-y)} \quad | \quad 2\sin(x - y)\cos(x-y) = \sin\Big(2(x-y)\Big) \\ \mathbf{F_x} &=& \mathbf{\dfrac{5 + x^2+x\sin\Big(2(x-y)\Big)}{\cos^2(x-y)}} \\ \hline \end{array} }$$

$$\small{ \begin{array}{|rcll|} \hline \mathbf{F_y = \ ?} \\ \hline F_y &=& \dfrac{d }{dy}(5 + x^2)\tan(x - y)- \dfrac{d}{dy}y \quad | \quad \dfrac{d}{dy}y = 1 \\ F_y &=& \dfrac{d }{dy}(5 + x^2)\tan(x - y)-1 \\ F_y &=& (5 + x^2)\dfrac{d }{dy}\tan(x - y)-1 \quad | \quad \dfrac{d }{dx}\tan(x - y) = \dfrac{1}{\cos^2(x-y)}*(-1) \\ F_y &=& (5 + x^2)\dfrac{1}{\cos^2(x-y)}*(-1) -1 \\ F_y &=& -\left( \dfrac{(5 + x^2)}{\cos^2(x-y)} +1 \right) \\ F_y &=& -\left( \dfrac{5 + x^2+\cos^2(x-y)}{\cos^2(x-y)} \right) \quad | \quad \cos^2(x-y)=\dfrac{1+ \cos\Big(2(x-y)\Big)}{2} \\ F_y &=& -\left( \dfrac{5 + x^2+\dfrac{1+ \cos\Big(2(x-y)\Big)}{2}}{\cos^2(x-y)} \right) \\ F_y &=& -\left( \dfrac{10 + 2x^2+1+ \cos\Big(2(x-y)\Big)}{2\cos^2(x-y)} \right) \\ \mathbf{F_y} &=& \mathbf{ -\left( \dfrac{11 + 2x^2+\cos\Big(2(x-y)\Big)}{2\cos^2(x-y)} \right)} \\ \hline \end{array} }$$

$$\begin{array}{|rcll|} \hline y'(x) &=& -\dfrac{F_x}{F_y} \\\\ y'(x) &=& -\dfrac{\dfrac{5 + x^2+x\sin\Big(2(x-y)\Big)}{\cos^2(x-y)}} {-\left( \dfrac{11 + 2x^2+\cos\Big(2(x-y)\Big)}{2\cos^2(x-y)} \right)} \\\\ y'(x) &=& \dfrac{\dfrac{5 + x^2+x\sin\Big(2(x-y)\Big)}{\cos^2(x-y)}} {\dfrac{11 + 2x^2+\cos\Big(2(x-y)\Big)}{2\cos^2(x-y)} } \\\\ y'(x) &=& \dfrac{ 2\left( 5 + x^2+ x\sin \Big(2(x-y)\Big) \right) } { 11 + 2x^2 +\cos\Big(2(x-y)\Big)} \\\\ \mathbf{y'(x)} &=& \mathbf{\dfrac{ 2\Big( 5 + x^2+ x\sin(2x-2y) \Big) } { 11 + 2x^2 +\cos(2x-2y)} } \\ \hline \end{array}$$ Feb 26, 2020