Chain
f(x) = (2x+1)²
Find f'(x)
2(2x+1) • 2
OR
4(2x+1)
OR
8x+4
h(x) = (x²-4)(3x+2)
Find h'(x)
(3x+2)(2x)+(x²-4)(3)
OR
9x²+4x-12
f(x) = 3x
find f'(x)
3
Limit of f(x) as x approaches 3
f(x) = (x2-6x+9) / (x-3)
h(x) = (sin(x) + cos(x))2 / (3x3-4x+2x2-1)
h'(x)
((3x3-4x+2x2-1)(2)(cos2(x)-sin2(x)) - (9x2-4+4x)(sin(x)+cos(x))2) / (3x3-4x+2x2-1)2
f(x) = 1 / (x²+9x)²
Find f'(x)
-2(2x+9) / (x²+9x)³
This was the in class example on 10/01/25!
h(x) = (x⁴+2x²-1) / (3x²-4x+1)
Find h'(x)
((3x²-4x+1)(4x³+4x) - (x⁴+2x²-1)(6x-4) ) / (3x²-4x+1)²
f(x) = sin(x)
Find f'(x)cos(x)
Limit of f(x) as x approaches 4
f(x) = (x2-16)/(x-4)
8
h(x)= sin(x)cos(x)
Find h'(x)
-sin2x + cos2x
f(x) = (sin(x))² + 3x
Find f'(x)
2(sin(x))+cos(x)+3
h(x) = sin(x) / cos(x)
Find h'(x)
sec²x
f(x) = cos(x)
Find f'(x)
-sin(x)
Limit of f(x) as x approaches 2
f(x) = x-2/x2-4
1/4
As x approaches 0 of f(x), f(x) approaches what number
f(x) = -1/|x|
Negative infinity
f(x) = (3x³-2x+5x²-3) ^ (2/3)
2(9x²+10x-2) / 3(3x³ + 5x² -2x -3)
h(x) = (x²-2x+3)² • (3x+2)³
Find h'(x)
(3x²+2)³•(2(x²-2x+3)(2x-2)) + (x²-2x+3)²•(3(3x+2)²(3))
f(x) = tan(x)
f'(x)
sec2(x)
Limit of f(x) as x approaches 3
f(x) = (x2-3x) / (√(x+6) -3)
18
f(x) = tan(x)cos(x)
Find f'(x)
f'(x) = cos(x)
f(x) = ((x³+2x²+5x-3) ^(3/2)) / ((3x+2)^2)
(3x+2)²•(3/2)•(√(x³+2x²+5x-3)•(3x²+4x+5) + ((x³+2x²+5x-3)^(3/2) )•(2)•(3x+2)•(3)
h(x) = (x2+3x-1)5 / (x3+6x+2)3
Find h'(x)
( (x3+6x+2)3 * 5(x2+3x-1)4*(2x+3) - (x2+3x-1)5 *3(x3+6x+2)2*(3x2+6) ) / (x3+6x+2)6
f(x) = sec(x)
f'(x)
sec(x)tan(x)
Limit of f(x) as x approaches 0
f(x) = (x³+4x²+5x-10) / (x-1)
16
f(x) = e5sinx
f'(x)
f''(x)
f'(x) = 5cosx*e5sinx
f''(x) = -5sin(x)*e5sinx + 25cos2(x)*e5sinx