TECHNIQUES OF DIFFERENTIATION
๐ณMTS201
*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.๐๐ฝ
*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 this๐ค
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=-1๐
*Written by:* Aluko AGY Dept.
*Edited:*©️ Flashpee_D_Wave๐
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