Chain rule, product rule, quotient rule of differenciation formulae example and it's solution

 

Chain rule

When differenciating an independent 

dy/dx =du/dx . dy/dx

 Product rule

If y = uv the product between u and v is  given by the following equation.

udv/dx + vdu/dx

Quotient rule

The quotient rule state that when ever

 y = u/v  the best way to differenciate is by applying the below equation.

(vdu - udv)/v²

Examples of CHAIN RULE and it's solution.

Eg1)

y=sin(2x² - 4 )

Eg2)

y= tan(9x)

Eg3) y=x³

         Solution to eg 1.

u=(2x³-4)  du/dx = 6x²

y=sinu ,  dy/dx= cosu

d/dx=dy/dx * du/dx

= Cosu*6x²

=6x²cosu ⇔ u=(2x³-4)

=6x²cos(2x³-4)

       Solution to eg 2.

y=tan(9x) , u= 9x now y= tan(u)

du/dx=9 

dy/dx=sec(u)⇔ dy/dx*du/dx

sec(u)*9 therefore is the same thing as 9sec(u) but (u) is equal to 9x, by recalling that, now 

y=tan(9x) is equivalent to 9sec(9x).

   Solution to eg 3.

y=x³, u= x³, du/dx=3x²

Now y=u , dy/dx=3x²

dy/du*du/dx=3x²*3x² ⁼ 9x²

      PRODUCT RULE examples and solutions

eg

y=lnx x²

y=eˣ cosx

      Soln eg1)

udv/dx+vdu/dx = u=lnx,  v=x² , du/dx=1/x , dv/dx= 2x

lnx2x+x²1/x = 2xlnx + x²/x

      soln eg2)

u= eˣ, v= cosx,  du= eˣ, dv= -sinx

eˣ-sinx + cosxeˣ= -sinxeˣ+cosxeˣ⁰

QUOTIENT RULE and it's example.

 y=4x³/x²    u= 4x³,  v=x² , du= 12x², dv= 2x

(X²12x² - 4x³2x)/(x²)² but (x²)(x²)= (x²)²

Therefore (12x²-4x³2x)/(x²)

11x²- 4x³2x