Question

EXAMPLE2 Differentiate
(i) log sin x2​

Answer 1

Answer:

Using product rule,

From logsinx²,

Consider sinx² first and diff separately using chain's rule

Let x²=a

>da/dx= 2x

So since 2x is our new value for a

Let's represent sina=b ,since x²=a,therefore we now have sina right?

Follow me

So we also diff sina=b by >db/da=cosa ,when we diff sin we have cos..

Now we have cos

And by chain's rule,db/da * da/dx= db/dx, so we multiple db/da and da/ dx to arrive at db/dx....therefore we have 2x*cosa= 2xcosa when a=x², now we have 2xcosx² = db/dx and now so all we have is log or Ln2xcosx² from the original equation which was lnsinx 2 and we know that lnx or logx= 1/x,therefore, let all of that Ln2xcosx²=x, so Dy/dx of lnx =1/x now when our x is Ln2xcosx², Dy/dx=1/Ln2xcosx² and our final answer is 1/Ln2xcosx²

Step-by-step explanation: