Asked by Foster
Differentiate from first principle:
i) y=x^2-3x
ii) y=5sinx
i) y=x^2-3x
ii) y=5sinx
Answers
Answered by
bobpursley
I will do one:
y'= Lim ((x+dx)^2-3(x+dx)-x^2+3x)/dx as dx>>zero
= Lim (x^2+2cdx+dx^2-3x-3dx-x^2+3x)/dx
=lim( 2x+dx-3) as dx>0
= 2x-3
y'= Lim ((x+dx)^2-3(x+dx)-x^2+3x)/dx as dx>>zero
= Lim (x^2+2cdx+dx^2-3x-3dx-x^2+3x)/dx
=lim( 2x+dx-3) as dx>0
= 2x-3
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