AboutAbe Mantell Expertise Hello,
I am a college professor of mathematics and regularly teach all levels
from elementary mathematics through differential equations, and would
be happy to assist anyone with such questions!
Experience Over 15 years teaching at the college level.
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)
