3.1
The Chain Rule
The Chain Rule
y = (2x + 3)^4
y'=?
8(2x+3)^3
x^2 + y^2 = 25
{dy}/{dx}=?
-x/y
f(x) = 5 - 2x^3
(f^-1)^'(7)=?
-1/6
f(x) = \sin^{-1}(2x)
f'(x)= ?
\frac{2}{\sqrt{1 - 4x^2}}
f(x) = \text{arccos}(x) + \text{arcsin}(x)
f'(1) = ?
0
f(x) = 3x^5 - 4x^3 + 5x
f`'` `'`(x) = ?
60x^3 - 24x
\frac{d}{dx}[\ln(3x^2 + 2x)]=?
\frac{6x + 2}{3x^2 + 2x}
x^2 + 2y^5 = 10xy
{dy}/{dx}=?
\frac{5y - x}{5y^4 - 5x}
f(x) = \sin(x)
(f^-1)^'(1/2)=?
{2\sqrt{3}}/3
g(x) = \text{arctan}(3x^2 + 4)
g'(x) = ?
\frac{6x}{1 + (3x^2 + 4)^2}
f(x) = x \cdot g(h(x))
g(4)=2 , g'(4)=3 ,
h(3)=4, h'(3)=–2
f'(3)=?
–16
y= x e^x
{d^2y}/{dx^2} = ?
xe^x + 2e^x
y = e^{x^2 + 3x}
\frac{dy}{dx}=?
(2x + 3) e^{x^2 + 3x}
xy^2 + y^3 = 8x
{dy}/{dx}=?
\frac{8 - y^2}{2xy + 3y^2}
f(x) = \sqrt{x-4}
(f^-1)^'(2)=?
\frac{d}{dx}[\csc^{–1}(x^6)]=?
-{6x^5}/{|x^6| \sqrt{x^12 - 1}}
\text{Let}\ f(x) = x^3 + 3x + 1,
g(x)=f^{-1}(x), \ \text{and}\ g(–3)= –1
\text{Find}\ g'(–3)
1/6
y= sin^2(x)
{d^2y}/{dx^2} = ?
2 \cos^2(x) - 2 \sin^2(x)
g(x) = \sqrt{5x^2 + 3}
g'(x)=?
\frac{5x}{\sqrt{5x^2 + 3}}
\sin(xy) = x+y
{dy}/{dx}=?
\frac{1 - y \cos(xy)}{x \cos(xy) - 1}
f(x) = {x+6}/{x-2}
(f^-1)^'(3)=?
-2
y= \cos^{–1}(x^3 + 2x)
dy/dx=?
dy/dx = - \frac{3x^2 + 2}{\sqrt{1 - (x^3 + 2x)^2}}
h(x) = \sqrt{x}
2 h`'``'`(4)=?
- 1/16
x^2 + y^2 = 9
{d^2y}/{dx^2}=?
-(x^2 + y^2)/y^3 = -9/{y^3}
f(x) = \cos^2(4x)
f'(x)=?
-8 \cos(4x) \cdot \sin(4x)
Find the equation of the tangent line.
x^2 + 7y^2 = 8y^3 \ \text{at} (-6,2)
(y - 2) = -3/17 (x + 6)
f(x) = \ln(2x)
(f^-1)^'(1)=?
e/2
Find the slope of the tangent line when x = 1
y = \text{arccsc}(5x)
\sqrt{6}/{12}
Find the equation of the tangent line when \theta = \pi/4
f(\theta)= \tan(\theta) \sin(\theta)
(y - \sqrt{2}/2) = {3 \sqrt{2}}/2 ( x - \pi/4)
\sin(x+y) = 2x
{d^2y}/{dx^2}=?
4 \sec^2(x+y) \tan(x+y)