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Calculus Examples
y=|x|√2-x2 , (1,1)
Step 1
Step 1.1
Use n√ax=axn to rewrite √2-x2 as (2-x2)12.
ddx[|x|(2-x2)12]
Step 1.2
Differentiate using the Quotient Rule which states that ddx[f(x)g(x)] is g(x)ddx[f(x)]-f(x)ddx[g(x)]g(x)2 where f(x)=|x| and g(x)=(2-x2)12.
(2-x2)12ddx[|x|]-|x|ddx[(2-x2)12]((2-x2)12)2
Step 1.3
Multiply the exponents in ((2-x2)12)2.
Step 1.3.1
Apply the power rule and multiply exponents, (am)n=amn.
(2-x2)12ddx[|x|]-|x|ddx[(2-x2)12](2-x2)12⋅2
Step 1.3.2
Cancel the common factor of 2.
Step 1.3.2.1
Cancel the common factor.
(2-x2)12ddx[|x|]-|x|ddx[(2-x2)12](2-x2)12⋅2
Step 1.3.2.2
Rewrite the expression.
(2-x2)12ddx[|x|]-|x|ddx[(2-x2)12](2-x2)1
(2-x2)12ddx[|x|]-|x|ddx[(2-x2)12](2-x2)1
(2-x2)12ddx[|x|]-|x|ddx[(2-x2)12](2-x2)1
Step 1.4
Simplify.
(2-x2)12ddx[|x|]-|x|ddx[(2-x2)12]2-x2
Step 1.5
The derivative of |x| with respect to x is x|x|.
(2-x2)12x|x|-|x|ddx[(2-x2)12]2-x2
Step 1.6
Combine (2-x2)12 and x|x|.
(2-x2)12x|x|-|x|ddx[(2-x2)12]2-x2
Step 1.7
Multiply (2-x2)12x|x|-|x|ddx[(2-x2)12]2-x2 by |x||x|.
|x||x|⋅(2-x2)12x|x|-|x|ddx[(2-x2)12]2-x2
Step 1.8
Combine.
|x|((2-x2)12x|x|-|x|ddx[(2-x2)12])|x|(2-x2)
Step 1.9
Apply the distributive property.
|x|(2-x2)12x|x|+|x|(-|x|ddx[(2-x2)12])|x|(2-x2)
Step 1.10
Cancel the common factor of |x|.
Step 1.10.1
Cancel the common factor.
|x|(2-x2)12x|x|+|x|(-|x|ddx[(2-x2)12])|x|(2-x2)
Step 1.10.2
Rewrite the expression.
(2-x2)12x+|x|(-|x|ddx[(2-x2)12])|x|(2-x2)
(2-x2)12x+|x|(-|x|ddx[(2-x2)12])|x|(2-x2)
Step 1.11
To multiply absolute values, multiply the terms inside each absolute value.
(2-x2)12x-|x⋅x|ddx[(2-x2)12]|x|(2-x2)
Step 1.12
Raise x to the power of 1.
(2-x2)12x-|x1x|ddx[(2-x2)12]|x|(2-x2)
Step 1.13
Raise x to the power of 1.
(2-x2)12x-|x1x1|ddx[(2-x2)12]|x|(2-x2)
Step 1.14
Use the power rule aman=am+n to combine exponents.
(2-x2)12x-|x1+1|ddx[(2-x2)12]|x|(2-x2)
Step 1.15
Add 1 and 1.
(2-x2)12x-|x2|ddx[(2-x2)12]|x|(2-x2)
Step 1.16
Differentiate using the chain rule, which states that ddx[f(g(x))] is f′(g(x))g′(x) where f(x)=x12 and g(x)=2-x2.
Step 1.16.1
To apply the Chain Rule, set u as 2-x2.
(2-x2)12x-|x2|(ddu[u12]ddx[2-x2])|x|(2-x2)
Step 1.16.2
Differentiate using the Power Rule which states that ddu[un] is nun-1 where n=12.
(2-x2)12x-|x2|(12u12-1ddx[2-x2])|x|(2-x2)
Step 1.16.3
Replace all occurrences of u with 2-x2.
(2-x2)12x-|x2|(12(2-x2)12-1ddx[2-x2])|x|(2-x2)
(2-x2)12x-|x2|(12(2-x2)12-1ddx[2-x2])|x|(2-x2)
Step 1.17
To write -1 as a fraction with a common denominator, multiply by 22.
(2-x2)12x-|x2|(12(2-x2)12-1⋅22ddx[2-x2])|x|(2-x2)
Step 1.18
Combine -1 and 22.
(2-x2)12x-|x2|(12(2-x2)12+-1⋅22ddx[2-x2])|x|(2-x2)
Step 1.19
Combine the numerators over the common denominator.
(2-x2)12x-|x2|(12(2-x2)1-1⋅22ddx[2-x2])|x|(2-x2)
Step 1.20
Simplify the numerator.
Step 1.20.1
Multiply -1 by 2.
(2-x2)12x-|x2|(12(2-x2)1-22ddx[2-x2])|x|(2-x2)
Step 1.20.2
Subtract 2 from 1.
(2-x2)12x-|x2|(12(2-x2)-12ddx[2-x2])|x|(2-x2)
(2-x2)12x-|x2|(12(2-x2)-12ddx[2-x2])|x|(2-x2)
Step 1.21
Combine fractions.
Step 1.21.1
Move the negative in front of the fraction.
(2-x2)12x-|x2|(12(2-x2)-12ddx[2-x2])|x|(2-x2)
Step 1.21.2
Combine 12 and (2-x2)-12.
(2-x2)12x-|x2|((2-x2)-122ddx[2-x2])|x|(2-x2)
Step 1.21.3
Move (2-x2)-12 to the denominator using the negative exponent rule b-n=1bn.
(2-x2)12x-|x2|(12(2-x2)12ddx[2-x2])|x|(2-x2)
Step 1.21.4
Combine 12(2-x2)12 and |x2|.
(2-x2)12x-|x2|2(2-x2)12ddx[2-x2]|x|(2-x2)
(2-x2)12x-|x2|2(2-x2)12ddx[2-x2]|x|(2-x2)
Step 1.22
By the Sum Rule, the derivative of 2-x2 with respect to x is ddx[2]+ddx[-x2].
(2-x2)12x-|x2|2(2-x2)12(ddx[2]+ddx[-x2])|x|(2-x2)
Step 1.23
Since 2 is constant with respect to x, the derivative of 2 with respect to x is 0.
(2-x2)12x-|x2|2(2-x2)12(0+ddx[-x2])|x|(2-x2)
Step 1.24
Add 0 and ddx[-x2].
(2-x2)12x-|x2|2(2-x2)12ddx[-x2]|x|(2-x2)
Step 1.25
Since -1 is constant with respect to x, the derivative of -x2 with respect to x is -ddx[x2].
(2-x2)12x-|x2|2(2-x2)12(-ddx[x2])|x|(2-x2)
Step 1.26
Multiply.
Step 1.26.1
Multiply -1 by -1.
(2-x2)12x+1|x2|2(2-x2)12ddx[x2]|x|(2-x2)
Step 1.26.2
Multiply |x2|2(2-x2)12 by 1.
(2-x2)12x+|x2|2(2-x2)12ddx[x2]|x|(2-x2)
(2-x2)12x+|x2|2(2-x2)12ddx[x2]|x|(2-x2)
Step 1.27
Differentiate using the Power Rule which states that ddx[xn] is nxn-1 where n=2.
(2-x2)12x+|x2|2(2-x2)12(2x)|x|(2-x2)
Step 1.28
Simplify terms.
Step 1.28.1
Combine 2 and |x2|2(2-x2)12.
(2-x2)12x+2|x2|2(2-x2)12x|x|(2-x2)
Step 1.28.2
Combine 2|x2|2(2-x2)12 and x.
(2-x2)12x+2|x2|x2(2-x2)12|x|(2-x2)
Step 1.28.3
Cancel the common factor.
(2-x2)12x+2|x2|x2(2-x2)12|x|(2-x2)
Step 1.28.4
Rewrite the expression.
(2-x2)12x+|x2|x(2-x2)12|x|(2-x2)
Step 1.28.5
Reorder 2 and -x2.
x(-x2+2)12+x|x2|(-x2+2)12|x|(2-x2)
x(-x2+2)12+x|x2|(-x2+2)12|x|(2-x2)
Step 1.29
To write x(-x2+2)12 as a fraction with a common denominator, multiply by (-x2+2)12(-x2+2)12.
x(-x2+2)12(-x2+2)12(-x2+2)12+x|x2|(-x2+2)12|x|(2-x2)
Step 1.30
Combine the numerators over the common denominator.
x(-x2+2)12(-x2+2)12+x|x2|(-x2+2)12|x|(2-x2)
Step 1.31
Multiply (-x2+2)12 by (-x2+2)12 by adding the exponents.
Step 1.31.1
Move (-x2+2)12.
x((-x2+2)12(-x2+2)12)+x|x2|(-x2+2)12|x|(2-x2)
Step 1.31.2
Use the power rule aman=am+n to combine exponents.
x(-x2+2)12+12+x|x2|(-x2+2)12|x|(2-x2)
Step 1.31.3
Combine the numerators over the common denominator.
x(-x2+2)1+12+x|x2|(-x2+2)12|x|(2-x2)
Step 1.31.4
Add 1 and 1.
x(-x2+2)22+x|x2|(-x2+2)12|x|(2-x2)
Step 1.31.5
Divide 2 by 2.
x(-x2+2)1+x|x2|(-x2+2)12|x|(2-x2)
x(-x2+2)1+x|x2|(-x2+2)12|x|(2-x2)
Step 1.32
Simplify x(-x2+2)1.
x(-x2+2)+x|x2|(-x2+2)12|x|(2-x2)
Step 1.33
Rewrite x(-x2+2)+x|x2|(-x2+2)12|x|(2-x2) as a product.
x(-x2+2)+x|x2|(-x2+2)12⋅1|x|(2-x2)
Step 1.34
Multiply x(-x2+2)+x|x2|(-x2+2)12 by 1|x|(2-x2).
x(-x2+2)+x|x2|(-x2+2)12(|x|(2-x2))
Step 1.35
Reorder terms.
x(-x2+2)+x|x2|(-x2+2)12(|x|(-x2+2))
Step 1.36
Multiply (-x2+2)12 by -x2+2 by adding the exponents.
Step 1.36.1
Move -x2+2.
x(-x2+2)+x|x2|(-x2+2)(-x2+2)12|x|
Step 1.36.2
Multiply -x2+2 by (-x2+2)12.
Step 1.36.2.1
Raise -x2+2 to the power of 1.
x(-x2+2)+x|x2|(-x2+2)1(-x2+2)12|x|
Step 1.36.2.2
Use the power rule aman=am+n to combine exponents.
x(-x2+2)+x|x2|(-x2+2)1+12|x|
x(-x2+2)+x|x2|(-x2+2)1+12|x|
Step 1.36.3
Write 1 as a fraction with a common denominator.
x(-x2+2)+x|x2|(-x2+2)22+12|x|
Step 1.36.4
Combine the numerators over the common denominator.
x(-x2+2)+x|x2|(-x2+2)2+12|x|
Step 1.36.5
Add 2 and 1.
x(-x2+2)+x|x2|(-x2+2)32|x|
x(-x2+2)+x|x2|(-x2+2)32|x|
Step 1.37
Simplify.
Step 1.37.1
Apply the distributive property.
x(-x2)+x⋅2+x|x2|(-x2+2)32|x|
Step 1.37.2
Simplify the numerator.
Step 1.37.2.1
Simplify each term.
Step 1.37.2.1.1
Rewrite using the commutative property of multiplication.
-x⋅x2+x⋅2+x|x2|(-x2+2)32|x|
Step 1.37.2.1.2
Multiply x by x2 by adding the exponents.
Step 1.37.2.1.2.1
Move x2.
-(x2x)+x⋅2+x|x2|(-x2+2)32|x|
Step 1.37.2.1.2.2
Multiply x2 by x.
Step 1.37.2.1.2.2.1
Raise x to the power of 1.
-(x2x1)+x⋅2+x|x2|(-x2+2)32|x|
Step 1.37.2.1.2.2.2
Use the power rule aman=am+n to combine exponents.
-x2+1+x⋅2+x|x2|(-x2+2)32|x|
-x2+1+x⋅2+x|x2|(-x2+2)32|x|
Step 1.37.2.1.2.3
Add 2 and 1.
-x3+x⋅2+x|x2|(-x2+2)32|x|
-x3+x⋅2+x|x2|(-x2+2)32|x|
Step 1.37.2.1.3
Move 2 to the left of x.
-x3+2⋅x+x|x2|(-x2+2)32|x|
Step 1.37.2.1.4
Remove non-negative terms from the absolute value.
-x3+2x+x⋅x2(-x2+2)32|x|
Step 1.37.2.1.5
Multiply x by x2 by adding the exponents.
Step 1.37.2.1.5.1
Multiply x by x2.
Step 1.37.2.1.5.1.1
Raise x to the power of 1.
-x3+2x+x1x2(-x2+2)32|x|
Step 1.37.2.1.5.1.2
Use the power rule aman=am+n to combine exponents.
-x3+2x+x1+2(-x2+2)32|x|
-x3+2x+x1+2(-x2+2)32|x|
Step 1.37.2.1.5.2
Add 1 and 2.
-x3+2x+x3(-x2+2)32|x|
-x3+2x+x3(-x2+2)32|x|
-x3+2x+x3(-x2+2)32|x|
Step 1.37.2.2
Combine the opposite terms in -x3+2x+x3.
Step 1.37.2.2.1
Add -x3 and x3.
2x+0(-x2+2)32|x|
Step 1.37.2.2.2
Add 2x and 0.
2x(-x2+2)32|x|
2x(-x2+2)32|x|
2x(-x2+2)32|x|
2x(-x2+2)32|x|
Step 1.38
Evaluate the derivative at x=1.
2(1)(-(1)2+2)32|1|
Step 1.39
Simplify.
Step 1.39.1
Multiply 2 by 1.
2(-12+2)32|1|
Step 1.39.2
Simplify the denominator.
Step 1.39.2.1
Simplify each term.
Step 1.39.2.1.1
One to any power is one.
2(-1⋅1+2)32|1|
Step 1.39.2.1.2
Multiply -1 by 1.
2(-1+2)32|1|
2(-1+2)32|1|
Step 1.39.2.2
Add -1 and 2.
2132|1|
Step 1.39.2.3
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.
2132⋅1
Step 1.39.2.4
One to any power is one.
21⋅1
21⋅1
Step 1.39.3
Simplify the expression.
Step 1.39.3.1
Multiply 1 by 1.
21
Step 1.39.3.2
Divide 2 by 1.
2
2
2
2
Step 2
Step 2.1
Use the slope 2 and a given point (1,1) 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-(1)=2⋅(x-(1))
Step 2.2
Simplify the equation and keep it in point-slope form.
y-1=2⋅(x-1)
Step 2.3
Solve for y.
Step 2.3.1
Simplify 2⋅(x-1).
Step 2.3.1.1
Rewrite.
y-1=0+0+2⋅(x-1)
Step 2.3.1.2
Simplify by adding zeros.
y-1=2⋅(x-1)
Step 2.3.1.3
Apply the distributive property.
y-1=2x+2⋅-1
Step 2.3.1.4
Multiply 2 by -1.
y-1=2x-2
y-1=2x-2
Step 2.3.2
Move all terms not containing y to the right side of the equation.
Step 2.3.2.1
Add 1 to both sides of the equation.
y=2x-2+1
Step 2.3.2.2
Add -2 and 1.
y=2x-1
y=2x-1
y=2x-1
y=2x-1
Step 3