tangent of f(x)=x^4-5x^2+2 at x=2
Find the tangent at x = 2
Let's perform the differentiation: ddx(x4 -5x2 + 2)
ⓘ The derivative of a sum is the sum of the derivatives: ddx(x4 -5x2 + 2) = ddx(x4) + ddx(-5x2) + 0
ddx(x4) + ddx(-5x2) + 0
ⓘ The derivative of x^n is n*x^(n-1): ddx(x4) = 4x3, ddx(x2) = 2x■Adding 0 changes nothing: ddx(-5x2) + 0 = ddx(-5x2)■A constant factor stays in front of the derivative, (k*u)' = k*u': ddx(-5x2) = -5·ddx(x2)
4x3 -5·2x
ⓘ Multiplications are done before additions and subtractions: 5·2 = 10
4x3 -10x
Compute f(2)
24 -5·22 + 2
ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: 24 = 2·2·2·2 = 16, 22 = 2·2 = 4
16 -5·4 + 2
ⓘ Multiplications are done before additions and subtractions: 5·4 = 20
16 -20 + 2
ⓘ Start from 16 and move back 20 on the number line, going below zero: 16 -20 = -4
-4 + 2
ⓘ Start from -4 and move forward 2 on the number line, getting closer to zero: -4 + 2 = -2
-2
The point of tangency is (2, -2)
Compute f'(2)
4·23 -10·2
ⓘ Powers are computed before multiplications, additions and subtractions. An integer power is a repeated product: 23 = 2·2·2 = 8■Multiplications are done before additions and subtractions: 10·2 = 20
4·8 -20
ⓘ Multiplications are done before additions and subtractions: 4·8 = 32
32 -20
ⓘ Start from 32 and move back 20 on the number line: 32 -20 = 12
12
The slope is f'(2) = 12
The tangent passes through this point with this slope: y = 12·(x -2) -2
Compute the y-intercept: b = -2 -12·2
-2 -12·2
ⓘ Multiplications are done before additions and subtractions: 12·2 = (10 + 2)·2 = 10·2 + 2·2 = 20 + 4 = 24
-2 -24
ⓘ Start from -2 and move back 24 more on the number line, away from zero: -2 -24 = -26
-26
▶Tangent line:   y = 12x -26