6 Differentiation Flashcards
how to differentiate y=x^n
dy/dx = nx^(n-1)
if f’(a) >0
when x=a, f is increasing
if f’(a) >0
when x=a, f is decreasing
if f(x)=3e^x -1/x show f(x) in increasing for x>0
f’(x) = 3e^x + 1/x^2
x>0 so e^x >0 and 1/x^2 >0
so 3e^x +1/x^2 >0
if f(x) =lnx what is f'(x)= and why
1/x
beacause its gradient is the reciprocal of y=e^x
dx/dy=
1/dy/dx
Ln3 what is dy/dx
1/0
Inverse of f(x) = e^x + 1
X=e^y + 1
Ln(x-1) = f^-1(x)
F(x) = ln3x -6
F^-1(x)=
Y+6 =ln(3x)
=1/3(e^x+6)
G(x) = 2e^(x-5) +3
G^-1(x) =
Range And domain
G^-1(x) = ln(1/2(x-3)) +5
Gs range= g(x) >3
G-1s domain = g-1 is x>3
Domain oF g(x) = xER
Range of g-1 — g-1 ER
F(x)= 5x^4 -ln(5x^4)
F’(x)=
F’(x)= 20x^3 - 4/x
Find stationary of y= 8x^2 -ln(x)
Dy/dx =0
X=+- 1/4
y= 8/16 -ln1/4 =1.886
-1/4 doses not exist because ln(-) is not possible
S.p (0.25, 1.886)
D2y/dx2 = 16 + 1/x^2
x=0.25
16 + 1/0.25^2 =32
32> 0 so minimum point
Product rule
Y can be split into u and v
U Du/dx
V. Dv/dx
Cross multiply
V du/dx + u dv/dx
differentiate y= x^1/2 lnx
u=x^1/2, du/dx = 1/2x^-1/2
v=lnx, du/dx=1/x
rearranged to x^-1/2(1/2lnx+1)
stationary points of x^2.e^x
find its type
dy/dx= e^x(2+x)x
dy/dx=0
e^x never equals 0
x=0
x=-2
x=0, y=0
x=-2, y=4e^-2
d2y/dx2 = e^x(4x+x^2+2)
x=0, d2y/dx2=2, 2>0 minimum
x=-2, d2y/dx2 =4e^-2, 4w^-2<0 maximum
quotient rule, how do you get there
product rule
du/dx =ydv/dx +vdy/dx
if y=u/v then you can rearange
dy/dx = (du/dx -(u/v)dv/dx)v
=(vdu/dx -udv/dx)v^2
when do you use the quotient rule
for fractions
y= (1+x^2) / (1+e^x) differentiate
y=u/v
2x+e^x(2x-1+x^2) /(1+e^x)^2 /(1+e^x)^2
how to find the gradient of y=sinx
x point you take is h
height of graph is thus sinh
and the x value is just h
gradient is sinh/h
make sure you’re in radians
f(x)= e^x(sinx) differentiate
e^x(sinx +cosx)
f(x)= tanx/(x^2 +1) =u/v
sec^2(x^2+1) -2xtanx /(x^2+1)^2
differentiate tanx
tanx =sinx/cosx
use quotint rule
cos^2x +sin^2x /cos^2x
1+tan^2x =sec2x
y=sin(x^2 -3) differentiat
is a function of a function
so you use chain rule du/dx x dv/dx
y=sinu, u= x^2 -3
dy/du= cosu, du/dx= 2x
dy/dx = 2xcos(x^2 -3)
differentiating sinx (all the others)
sinx > cosx > -sinx > -cosx > sinx
y= sqqrt(x^2 +1) differentiate
chain rule
dy/dx = x/sqqrt(x^2 +1)
chain rule
dydu x du/dx
y= ln(1+x^2) differentiate
chain rule
2x/(1+x^2)
extended chain rule
dy/dx = dy/du x du/dv x dv/dx
differentiate sqqrt(sinx^2)
ends up 1/2(sinx^2)^1/2 x cosx^2 x 2x
xcosx^2/sqqrt(sinx^2)
chain used it
somthing inside something
quotient used if
equation
product used if
used when its something x something
differentiate x^2e^x + 3lnx
e^x(2x+x^2) +3/x
y= x^2.e^-3x where x=1
find point
dy/dx= e^-3x(2x-3x^2)
x=1 e^-3(-1) =-e^-3
derivative of f(ax) is
af’(ax)
derivative of f(ax+b) is
af’(ax+b)
