Expert: Abe Mantell - 7/9/2008

Question
sir
how can i find the derivative of x raise to the power 5/2 by definition method? actually this problem can be easily solved by simple power rue but by definition i m not able to calculate (x - delta x) raise to the power 5/2.Can you please help me?

Answer
Hello Aliya (nice name)!

Let f(x)=x^(5/2), thus:
f'(x)=limit(h->0) [(x+h)^(5/2) - x^(5/2)]/h
now multiply by [(x+h)^(5/2) + x^(5/2)]/[(x+h)^(5/2) + x^(5/2)]
(i.e. the conjugate of (x+h)^(5/2) - x^(5/2))...which yields:
f'(x)=limit(h->0) [(x+h)^5 - x^5]/(h[(x+h)^(5/2) + x^(5/2)]),
since
(x+h)^5 - x^5 = x^5 + 5x^4*h + 10x^3*h^2 + 10x^2*h^3 + 5x*h^4 + h^5 - x^5
= 5x^4*h + 10x^3*h^2 + 10x^2*h^3 + 5x*h^4 + h^5, so we get:
f'(x)=limit(h->0)[5x^4*h+10x^3*h^2+10x^2*h^3+5x*h^4+h^5]/(h[(x+h)^(5/2)+x^(5/2)]),
factoring an "h" from the numerator and dividing by the "h" in the
denominator gives:
f'(x)=limit(h->0)[5x^4+10x^3*h+10x^2*h^2+5x*h^3+h^4]/([(x+h)^(5/2)+x^(5/2)]),
Now take the limit(h->0) to get:
f'(x)=[5x^4]/([x^(5/2)+x^(5/2)])=5x^4/[2x^(5/2)]
= (5/2)x^(3/2)

OK?

Abe