TECHNIQUES OF DIFFERENTIATION

๐Ÿ”ณMTS201

*TECHNIQUES OF DIFFERENTIATION*

sinx=cosx
cosx=-sinx
lnx=1/x
secx=secxtanx
tanx=sec²x
cosecx=-cotxcosecx
cotx=-cosecx

 For example, let's take a look at this question.
 *EXAMPLE* 
1. If y= x²sinx, let's find dy/dx...
 
 We should be familiar with this formula, _( *Udv/du +Vdu/dx)*_ 
 
 *Solution*
y=x²sinx
dy/dx= x²cosx+sinx2x
dy/dx=x²cosx+2xsinx✅
 

 Let's take a look at this question.๐Ÿ‘‡๐Ÿฝ

 *Example 2.*
 y=(x⁴+1)²
Find dy/dx...
 
 *Solution*
 dy/dx=2(x⁴+1)(4x³)
dy/dx=8x³(x⁴+1)✅
 
 There's what we all know as 
*IMPLICIT AND EXPLICIT DIFFERENTIATION*
 
 *Example of Implicit differentiation*
If y=x²+4xy⁴+7xy+8
dy/dx=....
 
 *Solution*
2x+4y⁴+16xy³ *dy/dx*+7y+7x *dy/dx*=0.

(16xy³+7x) *dy/dx*=-2x-4y⁴-7y

 *dy/dx*=-(2x+4y⁴+7y)/16xy³+7x✅
 

 *Example of Explicit Differentiation*
y=x²+2x-1
dy/dx=...
 
 *Solution*
 *dy/dx*=2x+2
...2(x+1)✅
 
 Now let's take a look at the

 *APPLICATION OF DIFFERENTIATION*
 
 *Example:*
 Find the equation of the tangent and normal to the curve 
y=2x³+3x²-2x-3 at the point x=1 and y=0...
 
 *Solution*
dy/dx=6x²+6x-2
Since x=1
m=6(1)²+(1)-2
m=12-2
=10
Equation on a straight line
y=mx+c
y=10x+c
0=10(1)+c
c=-10 
y=10x-10
y=10(x-1)✅
 
 ...to find equation of normal
Mm1=-1
m1=-1/M
  m1=-1/10
y-0=-1/10(x-1)
... therefore y=-x/10+1/10✅
 
 To find minimum and maximum point we differentiate twice...
I think we all remember this๐Ÿค”
 

 And we should also note that the positive integer greater than zero is *minimum* while the negative integer lesser than zero is *maximum*..
 
 *Example*:
 Consider the function
y=x³/3-x²/2-2x+5...
  *Solution*
dy/dx=x²-x-2
(x-2)(x+1)
...therefore x=2 or x=-1
To find the minimum and maximum point we differentiate twice...
d²y/dx²=2x-1
Since x=2
2(2)-1=3✅
3>0 therefore that's the *minimum* point
  2(-1)-1=-3✅
-3<0...we have the *maximum* point

  So, therefore the point is minimum at x=2 and maximum at x=-1๐Ÿ˜‡

 *Written by:* Aluko AGY Dept.

 *Edited:*©️ Flashpee_D_Wave๐Ÿ“š

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