Algebra
Calculus
Trigonometry
Matrix
Question
Function
Find the first partial derivative with respect to x
Find the first partial derivative with respect to c
∂x∂k=(c−x)22c2−2cx+x2
Evaluate
k=c×1−xx×(c×2−x)
Simplify
More Steps
Evaluate
c×1−xx×(c×2−x)
Use the commutative property to reorder the terms
c×1−xx×(2c−x)
Any expression multiplied by 1 remains the same
c−xx×(2c−x)
Multiply the terms
c−xx(2c−x)
k=c−xx(2c−x)
Find the first partial derivative by treating the variable c as a constant and differentiating with respect to x
∂x∂k=∂x∂(c−xx(2c−x))
Use differentiation rule ∂x∂(g(x)f(x))=(g(x))2∂x∂(f(x))×g(x)−f(x)×∂x∂(g(x))
∂x∂k=(c−x)2∂x∂(x(2c−x))(c−x)−x(2c−x)×∂x∂(c−x)
Evaluate
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Evaluate
∂x∂(x(2c−x))
Use differentiation rule ∂x∂(f(x)×g(x))=∂x∂(f(x))×g(x)+f(x)×∂x∂(g(x))
∂x∂(x)(2c−x)+x×∂x∂(2c−x)
Use ∂x∂xn=nxn−1 to find derivative
1×(2c−x)+x×∂x∂(2c−x)
Evaluate
2c−x+x×∂x∂(2c−x)
Evaluate
2c−x+x(−1)
Evaluate
2c−x−x
Subtract the terms
2c−2x
∂x∂k=(c−x)2(2c−2x)(c−x)−x(2c−x)×∂x∂(c−x)
Evaluate
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Evaluate
∂x∂(c−x)
Use differentiation rule ∂x∂(f(x)±g(x))=∂x∂(f(x))±∂x∂(g(x))
∂x∂(c)−∂x∂(x)
Use ∂x∂(c)=0 to find derivative
0−∂x∂(x)
Use ∂x∂xn=nxn−1 to find derivative
0−1
Removing 0 doesn't change the value,so remove it from the expression
−1
∂x∂k=(c−x)2(2c−2x)(c−x)−x(2c−x)(−1)
Evaluate
∂x∂k=(c−x)2(2c−2x)(c−x)−(−x(2c−x))
Solution
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Evaluate
(2c−2x)(c−x)−(−x(2c−x))
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
(2c−2x)(c−x)+x(2c−x)
Expand the expression
2c2−4cx+2x2+x(2c−x)
Expand the expression
2c2−4cx+2x2+2xc−x2
Add the terms
2c2−2cx+2x2−x2
Subtract the terms
2c2−2cx+x2
∂x∂k=(c−x)22c2−2cx+x2
Show Solution
Solve the equation
Solve for x
Solve for c
Solve for k
x=22c+k+4c2+k2x=22c+k−4c2+k2
Evaluate
k=c×1−xx×(c×2−x)
Simplify
More Steps
Evaluate
c×1−xx×(c×2−x)
Use the commutative property to reorder the terms
c×1−xx×(2c−x)
Any expression multiplied by 1 remains the same
c−xx×(2c−x)
Multiply the terms
c−xx(2c−x)
k=c−xx(2c−x)
Swap the sides of the equation
c−xx(2c−x)=k
Cross multiply
x(2c−x)=(c−x)k
Simplify the equation
x(2c−x)=k(c−x)
Calculate
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Evaluate
x(2c−x)
Apply the distributive property
x×2c−x×x
Use the commutative property to reorder the terms
2cx−x×x
Multiply the terms
2cx−x2
2cx−x2=k(c−x)
Calculate
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Evaluate
k(c−x)
Apply the distributive property
kc−kx
Use the commutative property to reorder the terms
ck−kx
2cx−x2=ck−kx
Move the expression to the left side
2cx−x2−(ck−kx)=0
If a negative sign or a subtraction symbol appears outside parentheses, remove the parentheses and change the sign of every term within the parentheses
2cx−x2−ck+kx=0
Collect like terms by calculating the sum or difference of their coefficients
(2c+k)x−x2−ck=0
Rewrite in standard form
−x2+(2c+k)x−ck=0
Multiply both sides
x2+(−2c−k)x+ck=0
Substitute a=1,b=−2c−k and c=ck into the quadratic formula x=2a−b±b2−4ac
x=22c+k±(−2c−k)2−4ck
Simplify the expression
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Evaluate
(−2c−k)2−4ck
Evaluate the power
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Evaluate
(−2c−k)2
A negative base raised to an even power equals a positive
(2c+k)2
Use (a+b)2=a2+2ab+b2 to expand the expression
(2c)2+2×2ck+k2
Calculate
4c2+4ck+k2
4c2+4ck+k2−4ck
Since two opposites add up to 0,remove them form the expression
4c2+k2
x=22c+k±4c2+k2
Solution
x=22c+k+4c2+k2x=22c+k−4c2+k2
Show Solution