I'm basic
y = 25
{dy}/{dx} = ?
0
f(x) = (x^2 + 1)^3
f'(1) = ?
f'(x) = 6x (x^2 + 1)^2
f'(1) = 24
f(x) = {x+6}/{x-2}
f'(1) = ?
f'(x) = {-8}/{(x-2)^2}
f'(1) = -8
f(x) = 3x^5 - 4x^3 + 5x
f`'` `'`(x) = ?
60x^3 - 24x
f(x) = (x^2 \cdot x^{-2/3})
f'(27) = ?
y = 3x
{dy}/{dx} = ?
3
g(x) = (2x + 3)^4
g'(-1) = ?
g'(x) = 8(2x+3)^3
g'(-1) = 8
f(x) = (x^2 + 2x)(4x - 5)
f'(x) = ?
f'(x) = 12x^2 + 6x - 10
y = sin(x)
{d^3y}/{dx^3} = ?
-cos(x)
Find the equation of the tangent line to the curve y = \sqrt{x} at x = 4
Hint: What does y equal when x = 4?
(y - 2) = 1/4 (x - 4)
Find the equation of the tangent line to the curve y = x^2 + 3x at x = 1
Hint: What does y equal when x = 1?
(y - 4) = 5 (x - 1)
g(x) = \sqrt{5x^2 + 3}
g'(1) = ?
g'(x) = \frac{5x}{\sqrt{5x^2 + 3}}
g'(1) = 5/3
f(x) = {3x^2 + 2}/{x^3 - 1}
f'(-1) = ?
f'(x) = {6x(x^3 - 1) - (3x^2 + 2)(3x^2)}/{(x^3 - 1)^2}
f'(-1) = -3/4
h(x) = \sqrt{x}
2 h`'``'`(4)=?
- 1/16
The velocity of an automobile starting from rest is
v(t) = {90t}/{4t + 10}
where v is measured in feet per second.
Find the acceleration at 5 seconds.
a(t) = {900}/{(4t + 10)^2}
a(5) = 1 ft/sec^2
f(x) = {5}/{2x^-4}
f'(2) = ?
f'(x) = 10x^3
f'(2) = 80
f(x) = \cos^2(4x)
f'(\pi/16) = ?
f'(x) = -8 \cos(4x) \cdot \sin(4x)
f'(\pi/16) = -4
f(\theta)= \tan(\theta) \cdot \sin(\theta)
f'(\pi / 4) = ?
f'(\theta) = \sec^2 x \sinx + \sinx = \sinx(\sec^2 x + 1)
f'(\pi/4) = 3/{\sqrt{2}} = {3\sqrt{2}}/2
h(x) = \root[3]{x^4}
h`'``'`(8)" = ?
h'(x) = 4/3 x^{1/3}
h''(x) = 4/{9 x^{2/3}} = 4/{9 \root[3]{x^2}}
h''(8) = 1/9
f(x) = x \cdot g(h(x))
g(4)=2, g'(4)=3, h(3)=4, h'(3)=-2
f'(3) = ?
–16
f(x) = {27 \root[3]{x^2}}/{3}
f'(8) = ?
f'(x) = 6/{x^{1/3}} = 6/{\root[3]{x}
f'(8) = 3
f(x) = 5 \root[3]{x^3 + 2x - 4}
f'(2) = ?
f'(x) = {15x^2 + 10}/{3 (x^3 + 2x - 4)^{2/3}
f'(2) = 70/12 = 35/6
f(x) = {\cos x}/{\sqrt{x}}
f'(\pi) = ?
f'(x) = {-\sqrt{x}\sin x - 1/{2\sqrt{x}} \cos x}/{x}
f'(\pi) = 1/{2\pi \sqrt{\pi}} = 1/{2 \sqrt{\pi^3}}
f(x) = (5x - 8)^3
f`'` `'`(2) = ?
f`'` `'`(x) = 150 (5x - 8)
f`'` `'`(2) = 300
Find the equation of the tangent line at the point (\pi/2, 1)
f(x) = {1 + \cos x}/{1 - \cos x}
f'(x) = {-2 \sin x}/{(1 - \cos x)^2}
f'(\pi/2) = -2
(y - 1) = -2 (x - \pi/2)