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Calculus Examples
2x2+3xy+4y2-5x+2y2x2+3xy+4y2−5x+2y
Step 1
Consider the limit definition of the derivative.
f′(x)=limh→0f(x+h)-f(x)h
Step 2
Step 2.1
Evaluate the function at x=x+h.
Step 2.1.1
Replace the variable x with x+h in the expression.
f(x+h)=2(x+h)2+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2
Simplify the result.
Step 2.1.2.1
Simplify each term.
Step 2.1.2.1.1
Rewrite (x+h)2 as (x+h)(x+h).
f(x+h)=2((x+h)(x+h))+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.2
Expand (x+h)(x+h) using the FOIL Method.
Step 2.1.2.1.2.1
Apply the distributive property.
f(x+h)=2(x(x+h)+h(x+h))+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.2.2
Apply the distributive property.
f(x+h)=2(x⋅x+xh+h(x+h))+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.2.3
Apply the distributive property.
f(x+h)=2(x⋅x+xh+hx+h⋅h)+3(x+h)⋅y+4y2-5(x+h)+2y
f(x+h)=2(x⋅x+xh+hx+h⋅h)+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.3
Simplify and combine like terms.
Step 2.1.2.1.3.1
Simplify each term.
Step 2.1.2.1.3.1.1
Multiply x by x.
f(x+h)=2(x2+xh+hx+h⋅h)+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.3.1.2
Multiply h by h.
f(x+h)=2(x2+xh+hx+h2)+3(x+h)⋅y+4y2-5(x+h)+2y
f(x+h)=2(x2+xh+hx+h2)+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.3.2
Add xh and hx.
Step 2.1.2.1.3.2.1
Reorder x and h.
f(x+h)=2(x2+hx+hx+h2)+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.3.2.2
Add hx and hx.
f(x+h)=2(x2+2hx+h2)+3(x+h)⋅y+4y2-5(x+h)+2y
f(x+h)=2(x2+2hx+h2)+3(x+h)⋅y+4y2-5(x+h)+2y
f(x+h)=2(x2+2hx+h2)+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.4
Apply the distributive property.
f(x+h)=2x2+2(2hx)+2h2+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.5
Multiply 2 by 2.
f(x+h)=2x2+4(hx)+2h2+3(x+h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.6
Apply the distributive property.
f(x+h)=2x2+4hx+2h2+(3x+3h)⋅y+4y2-5(x+h)+2y
Step 2.1.2.1.7
Apply the distributive property.
f(x+h)=2x2+4hx+2h2+3xy+3hy+4y2-5(x+h)+2y
Step 2.1.2.1.8
Apply the distributive property.
f(x+h)=2x2+4hx+2h2+3xy+3hy+4y2-5x-5h+2y
f(x+h)=2x2+4hx+2h2+3xy+3hy+4y2-5x-5h+2y
Step 2.1.2.2
The final answer is 2x2+4hx+2h2+3xy+3hy+4y2-5x-5h+2y.
2x2+4hx+2h2+3xy+3hy+4y2-5x-5h+2y
2x2+4hx+2h2+3xy+3hy+4y2-5x-5h+2y
2x2+4hx+2h2+3xy+3hy+4y2-5x-5h+2y
Step 2.2
Reorder.
Step 2.2.1
Move -5x.
2x2+4hx+2h2+3xy+3hy+4y2-5h-5x+2y
Step 2.2.2
Move 3xy.
2x2+4hx+2h2+3hy+3xy+4y2-5h-5x+2y
Step 2.2.3
Move 2x2.
4hx+2h2+3hy+2x2+3xy+4y2-5h-5x+2y
Step 2.2.4
Reorder 4hx and 2h2.
2h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y
2h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y
Step 2.3
Find the components of the definition.
f(x+h)=2h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y
f(x)=2x2+3xy+4y2-5x+2y
f(x+h)=2h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y
f(x)=2x2+3xy+4y2-5x+2y
Step 3
Plug in the components.
f′(x)=limh→02h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y-(2x2+3xy+4y2-5x+2y)h
Step 4
Step 4.1
Simplify the numerator.
Step 4.1.1
Apply the distributive property.
f′(x)=limh→02h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y-(2x2)-(3xy)-(4y2)-(-5x)-(2y)h
Step 4.1.2
Simplify.
Step 4.1.2.1
Multiply 2 by -1.
f′(x)=limh→02h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y-2x2-(3xy)-(4y2)-(-5x)-(2y)h
Step 4.1.2.2
Multiply 3 by -1.
f′(x)=limh→02h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y-2x2-3(xy)-(4y2)-(-5x)-(2y)h
Step 4.1.2.3
Multiply 4 by -1.
f′(x)=limh→02h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y-2x2-3(xy)-4y2-(-5x)-(2y)h
Step 4.1.2.4
Multiply -5 by -1.
f′(x)=limh→02h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y-2x2-3(xy)-4y2+5x-(2y)h
Step 4.1.2.5
Multiply 2 by -1.
f′(x)=limh→02h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y-2x2-3(xy)-4y2+5x-2yh
f′(x)=limh→02h2+4hx+3hy+2x2+3xy+4y2-5h-5x+2y-2x2-3(xy)-4y2+5x-2yh
Step 4.1.3
Subtract 2x2 from 2x2.
f′(x)=limh→02h2+4hx+3hy+3xy+4y2-5h-5x+2y+0-3xy-4y2+5x-2yh
Step 4.1.4
Add 2h2 and 0.
f′(x)=limh→02h2+4hx+3hy+3xy+4y2-5h-5x+2y-3xy-4y2+5x-2yh
Step 4.1.5
Subtract 3xy from 3xy.
f′(x)=limh→02h2+4hx+3hy+4y2-5h-5x+2y+0-4y2+5x-2yh
Step 4.1.6
Add 2h2 and 0.
f′(x)=limh→02h2+4hx+3hy+4y2-5h-5x+2y-4y2+5x-2yh
Step 4.1.7
Subtract 4y2 from 4y2.
f′(x)=limh→02h2+4hx+3hy-5h-5x+2y+0+5x-2yh
Step 4.1.8
Add 2h2 and 0.
f′(x)=limh→02h2+4hx+3hy-5h-5x+2y+5x-2yh
Step 4.1.9
Add -5x and 5x.
f′(x)=limh→02h2+4hx+3hy-5h+2y+0-2yh
Step 4.1.10
Add 2h2 and 0.
f′(x)=limh→02h2+4hx+3hy-5h+2y-2yh
Step 4.1.11
Subtract 2y from 2y.
f′(x)=limh→02h2+4hx+3hy-5h+0h
Step 4.1.12
Add 2h2+4hx+3hy-5h and 0.
f′(x)=limh→02h2+4hx+3hy-5hh
Step 4.1.13
Factor h out of 2h2+4hx+3hy-5h.
Step 4.1.13.1
Factor h out of 2h2.
f′(x)=limh→0h(2h)+4hx+3hy-5hh
Step 4.1.13.2
Factor h out of 4hx.
f′(x)=limh→0h(2h)+h(4x)+3hy-5hh
Step 4.1.13.3
Factor h out of 3hy.
f′(x)=limh→0h(2h)+h(4x)+h(3y)-5hh
Step 4.1.13.4
Factor h out of -5h.
f′(x)=limh→0h(2h)+h(4x)+h(3y)+h⋅-5h
Step 4.1.13.5
Factor h out of h(2h)+h(4x).
f′(x)=limh→0h(2h+4x)+h(3y)+h⋅-5h
Step 4.1.13.6
Factor h out of h(2h+4x)+h(3y).
f′(x)=limh→0h(2h+4x+3y)+h⋅-5h
Step 4.1.13.7
Factor h out of h(2h+4x+3y)+h⋅-5.
f′(x)=limh→0h(2h+4x+3y-5)h
f′(x)=limh→0h(2h+4x+3y-5)h
f′(x)=limh→0h(2h+4x+3y-5)h
Step 4.2
Reduce the expression by cancelling the common factors.
Step 4.2.1
Cancel the common factor of h.
Step 4.2.1.1
Cancel the common factor.
f′(x)=limh→0h(2h+4x+3y-5)h
Step 4.2.1.2
Divide 2h+4x+3y-5 by 1.
f′(x)=limh→02h+4x+3y-5
f′(x)=limh→02h+4x+3y-5
Step 4.2.2
Move 2h.
f′(x)=limh→04x+3y+2h-5
f′(x)=limh→04x+3y+2h-5
f′(x)=limh→04x+3y+2h-5
Step 5
Step 5.1
Split the limit using the Sum of Limits Rule on the limit as h approaches 0.
limh→04x+limh→03y+limh→02h-limh→05
Step 5.2
Evaluate the limit of 4x which is constant as h approaches 0.
4x+limh→03y+limh→02h-limh→05
Step 5.3
Evaluate the limit of 3y which is constant as h approaches 0.
4x+3y+limh→02h-limh→05
Step 5.4
Move the term 2 outside of the limit because it is constant with respect to h.
4x+3y+2limh→0h-limh→05
Step 5.5
Evaluate the limit of 5 which is constant as h approaches 0.
4x+3y+2limh→0h-1⋅5
4x+3y+2limh→0h-1⋅5
Step 6
Evaluate the limit of h by plugging in 0 for h.
4x+3y+2⋅0-1⋅5
Step 7
Step 7.1
Simplify each term.
Step 7.1.1
Multiply 2 by 0.
4x+3y+0-1⋅5
Step 7.1.2
Multiply -1 by 5.
4x+3y+0-5
4x+3y+0-5
Step 7.2
Add 4x+3y and 0.
4x+3y-5
4x+3y-5
Step 8
