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Calculus Examples
2xy-2x+y=-142xy−2x+y=−14 ; (2,-2)(2,−2)
Step 1
Step 1.1
Differentiate both sides of the equation.
ddx(2xy-2x+y)=ddx(-14)ddx(2xy−2x+y)=ddx(−14)
Step 1.2
Differentiate the left side of the equation.
Step 1.2.1
By the Sum Rule, the derivative of 2xy-2x+y2xy−2x+y with respect to xx is ddx[2xy]+ddx[-2x]+ddx[y]ddx[2xy]+ddx[−2x]+ddx[y].
ddx[2xy]+ddx[-2x]+ddx[y]ddx[2xy]+ddx[−2x]+ddx[y]
Step 1.2.2
Evaluate ddx[2xy]ddx[2xy].
Step 1.2.2.1
Since 22 is constant with respect to xx, the derivative of 2xy2xy with respect to xx is 2ddx[xy]2ddx[xy].
2ddx[xy]+ddx[-2x]+ddx[y]2ddx[xy]+ddx[−2x]+ddx[y]
Step 1.2.2.2
Differentiate using the Product Rule which states that ddx[f(x)g(x)]ddx[f(x)g(x)] is f(x)ddx[g(x)]+g(x)ddx[f(x)]f(x)ddx[g(x)]+g(x)ddx[f(x)] where f(x)=xf(x)=x and g(x)=yg(x)=y.
2(xddx[y]+yddx[x])+ddx[-2x]+ddx[y]2(xddx[y]+yddx[x])+ddx[−2x]+ddx[y]
Step 1.2.2.3
Rewrite ddx[y] as y′.
2(xy′+yddx[x])+ddx[-2x]+ddx[y]
Step 1.2.2.4
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
2(xy′+y⋅1)+ddx[-2x]+ddx[y]
Step 1.2.2.5
Multiply y by 1.
2(xy′+y)+ddx[-2x]+ddx[y]
2(xy′+y)+ddx[-2x]+ddx[y]
Step 1.2.3
Evaluate ddx[-2x].
Step 1.2.3.1
Since -2 is constant with respect to x, the derivative of -2x with respect to x is -2ddx[x].
2(xy′+y)-2ddx[x]+ddx[y]
Step 1.2.3.2
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=1.
2(xy′+y)-2⋅1+ddx[y]
Step 1.2.3.3
Multiply -2 by 1.
2(xy′+y)-2+ddx[y]
2(xy′+y)-2+ddx[y]
Step 1.2.4
Rewrite ddx[y] as y′.
2(xy′+y)-2+y′
Step 1.2.5
Apply the distributive property.
2xy′+2y-2+y′
2xy′+2y-2+y′
Step 1.3
Since -14 is constant with respect to x, the derivative of -14 with respect to x is 0.
0
Step 1.4
Reform the equation by setting the left side equal to the right side.
2xy′+2y-2+y′=0
Step 1.5
Solve for y′.
Step 1.5.1
Move all terms not containing y′ to the right side of the equation.
Step 1.5.1.1
Subtract 2y from both sides of the equation.
2xy′-2+y′=-2y
Step 1.5.1.2
Add 2 to both sides of the equation.
2xy′+y′=-2y+2
2xy′+y′=-2y+2
Step 1.5.2
Factor y′ out of 2xy′+y′.
Step 1.5.2.1
Factor y′ out of 2xy′.
y′(2x)+y′=-2y+2
Step 1.5.2.2
Raise y′ to the power of 1.
y′(2x)+y′=-2y+2
Step 1.5.2.3
Factor y′ out of y′1.
y′(2x)+y′⋅1=-2y+2
Step 1.5.2.4
Factor y′ out of y′(2x)+y′⋅1.
y′(2x+1)=-2y+2
y′(2x+1)=-2y+2
Step 1.5.3
Divide each term in y′(2x+1)=-2y+2 by 2x+1 and simplify.
Step 1.5.3.1
Divide each term in y′(2x+1)=-2y+2 by 2x+1.
y′(2x+1)2x+1=-2y2x+1+22x+1
Step 1.5.3.2
Simplify the left side.
Step 1.5.3.2.1
Cancel the common factor of 2x+1.
Step 1.5.3.2.1.1
Cancel the common factor.
y′(2x+1)2x+1=-2y2x+1+22x+1
Step 1.5.3.2.1.2
Divide y′ by 1.
y′=-2y2x+1+22x+1
y′=-2y2x+1+22x+1
y′=-2y2x+1+22x+1
Step 1.5.3.3
Simplify the right side.
Step 1.5.3.3.1
Combine the numerators over the common denominator.
y′=-2y+22x+1
Step 1.5.3.3.2
Factor 2 out of -2y+2.
Step 1.5.3.3.2.1
Factor 2 out of -2y.
y′=2(-y)+22x+1
Step 1.5.3.3.2.2
Factor 2 out of 2.
y′=2(-y)+2(1)2x+1
Step 1.5.3.3.2.3
Factor 2 out of 2(-y)+2(1).
y′=2(-y+1)2x+1
y′=2(-y+1)2x+1
Step 1.5.3.3.3
Factor -1 out of -y.
y′=2(-(y)+1)2x+1
Step 1.5.3.3.4
Rewrite 1 as -1(-1).
y′=2(-(y)-1(-1))2x+1
Step 1.5.3.3.5
Factor -1 out of -(y)-1(-1).
y′=2(-(y-1))2x+1
Step 1.5.3.3.6
Simplify the expression.
Step 1.5.3.3.6.1
Rewrite -(y-1) as -1(y-1).
y′=2(-1(y-1))2x+1
Step 1.5.3.3.6.2
Move the negative in front of the fraction.
y′=-2(y-1)2x+1
y′=-2(y-1)2x+1
y′=-2(y-1)2x+1
y′=-2(y-1)2x+1
y′=-2(y-1)2x+1
Step 1.6
Replace y′ with dydx.
dydx=-2(y-1)2x+1
Step 1.7
Evaluate at x=2 and y=-2.
Step 1.7.1
Replace the variable x with 2 in the expression.
-2(y-1)2(2)+1
Step 1.7.2
Replace the variable y with -2 in the expression.
-2((-2)-1)2(2)+1
Step 1.7.3
Subtract 1 from -2.
-2⋅-32(2)+1
Step 1.7.4
Simplify the denominator.
Step 1.7.4.1
Multiply 2 by 2.
-2⋅-34+1
Step 1.7.4.2
Add 4 and 1.
-2⋅-35
-2⋅-35
Step 1.7.5
Simplify the expression.
Step 1.7.5.1
Multiply 2 by -3.
--65
Step 1.7.5.2
Move the negative in front of the fraction.
--65
--65
Step 1.7.6
Multiply --65.
Step 1.7.6.1
Multiply -1 by -1.
1(65)
Step 1.7.6.2
Multiply 65 by 1.
65
65
65
65
Step 2
Step 2.1
Use the slope 65 and a given point (2,-2) to substitute for x1 and y1 in the point-slope form y-y1=m(x-x1), which is derived from the slope equation m=y2-y1x2-x1.
y-(-2)=65⋅(x-(2))
Step 2.2
Simplify the equation and keep it in point-slope form.
y+2=65⋅(x-2)
Step 2.3
Solve for y.
Step 2.3.1
Simplify 65⋅(x-2).
Step 2.3.1.1
Rewrite.
y+2=0+0+65⋅(x-2)
Step 2.3.1.2
Simplify by adding zeros.
y+2=65⋅(x-2)
Step 2.3.1.3
Apply the distributive property.
y+2=65x+65⋅-2
Step 2.3.1.4
Combine 65 and x.
y+2=6x5+65⋅-2
Step 2.3.1.5
Multiply 65⋅-2.
Step 2.3.1.5.1
Combine 65 and -2.
y+2=6x5+6⋅-25
Step 2.3.1.5.2
Multiply 6 by -2.
y+2=6x5+-125
y+2=6x5+-125
Step 2.3.1.6
Move the negative in front of the fraction.
y+2=6x5-125
y+2=6x5-125
Step 2.3.2
Move all terms not containing y to the right side of the equation.
Step 2.3.2.1
Subtract 2 from both sides of the equation.
y=6x5-125-2
Step 2.3.2.2
To write -2 as a fraction with a common denominator, multiply by 55.
y=6x5-125-2⋅55
Step 2.3.2.3
Combine -2 and 55.
y=6x5-125+-2⋅55
Step 2.3.2.4
Combine the numerators over the common denominator.
y=6x5+-12-2⋅55
Step 2.3.2.5
Simplify the numerator.
Step 2.3.2.5.1
Multiply -2 by 5.
y=6x5+-12-105
Step 2.3.2.5.2
Subtract 10 from -12.
y=6x5+-225
y=6x5+-225
Step 2.3.2.6
Move the negative in front of the fraction.
y=6x5-225
y=6x5-225
Step 2.3.3
Reorder terms.
y=65x-225
y=65x-225
y=65x-225
Step 3