(x4 + 1)5
20x3 (x4 + 1)4
3x3=4y2 + 2
(9x2)/(8y)
y=cos-1(-5x4)
(20x3) / (1 - 25x8)1/2
y=ln (4x5)
5/x
y= 2x2 - 4x - 2 at (2, -2)
Find the equation of the normal line.
y= -(1/4) x - 3/2
y = cos (5x4)
-20x3 sin (5x4)
-5xy2 + 5 = x
(-1 -5y2) / (10xy)
y=tan-1(4x5)
(20x4) / (16x10 + 1)
y=25x^4
5x3 * 25x^4 + 2 ln 2
y= -2 cos (2x) at (pi/4, 0)
Find the equation of the tangent line.
y = 4x - pi
(3x4 + 5)4 (5x3 + 1)
3x2 (3x4 + 5)3 (95x4 + 16x + 25)
(5y2 + 1)2 =4x3
(3x2) / (25y3 + 5y)
y=csc-1 (5x3)
-(15x2) / (|5x3| (25x6 - 1)1/2
y=log3(4x3)
3/ (x ln 3)
y= -x3 + x2
Find the points where the tangent line is horizontal.
(0,0), (2/3, 4/27)
y = sin3 (5x4)
60x3 sin2 (5x4) cos (5x4)
(3y3 + 2)2 =2x
1 / (27y5 + 18y2)
f(x) = x + 4, a=3
(f-1)' (a) = 1
y=5xx^3
y(3x2 ln x + x2)
x2 = 3y2 + 2
Find (d2y)/(dx2).
(d2 y)/(dx2) = (3y2 -x2) / (9y3)
y=cos (sin (2x5))
=-10x4 sin (sin (2x5)) cos (2x5)
3x + 5 = sec (2y2)
3 / (4y sec (2y2) tan (2y2))
f(x) = 5x - 3, a = -2
(f-1)' (a) = 1/5
y= ((4x3 + 3)4) / ((x +1)3 * (5x2 - 2)5)
[((4x3 + 3)4) / ((x +1)3 * (5x2 - 2)5) ] * [(48x2)/(4x3 + 3) - (3) / (x + 1) - (50x) / (5x2 - 2)]
5x2 -4y2 =3
Find (d2y)/(dx2).
(d2y) / (dx2) = (20y2 - 25x2) / (16y3)