Robert J. answered • 12/04/13

Certified High School AP Calculus and Physics Teacher

This is the derivative of x^5 at x = 0.5.

Lim h-->0 (((.5 + h)

^{5}) - ((.5)^{5}))/h= 5x^4 at x = 1/2

= 5/16 <==Answer

Jasmine J.

asked • 12/04/13Lim h-->0 (((.5 + h)^{5}) - ((.5)^{5}))/h

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Robert J. answered • 12/04/13

Certified High School AP Calculus and Physics Teacher

This is the derivative of x^5 at x = 0.5.

Lim h-->0 (((.5 + h)^{5}) - ((.5)^{5}))/h

= 5x^4 at x = 1/2

= 5/16 <==Answer

Steve S. answered • 12/05/13

Tutoring in Precalculus, Trig, and Differential Calculus

This problem is given before the students know about derivatives, so it must be solved with algebra.

To find (.5 + h)^5 either multiply four times, or use the binomial theorem with the help of Pascal's Triangle:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

(h + 1/2)^5 = h^5 + 5h^4(1/2) + 10h^3(1/2)^2 + 10h^2(1/2)^3 + 5h(1/2)^4 + (1/2)^5

Then (((.5 + h)^5) - ((.5)^5))/h

= (h^5 + 5h^4(1/2) + 10h^3(1/2)^2 + 10h^2(1/2)^3 + 5h(1/2)^4 + (1/2)^5 - (1/2)^5)/h

= h^4 + 5h^3(1/2) + 10h^2(1/2)^2 + 10h(1/2)^3 + 5(1/2)^4

Now as h —> 0 all of the terms with h in them go away and:

To find (.5 + h)^5 either multiply four times, or use the binomial theorem with the help of Pascal's Triangle:

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

(h + 1/2)^5 = h^5 + 5h^4(1/2) + 10h^3(1/2)^2 + 10h^2(1/2)^3 + 5h(1/2)^4 + (1/2)^5

Then (((.5 + h)^5) - ((.5)^5))/h

= (h^5 + 5h^4(1/2) + 10h^3(1/2)^2 + 10h^2(1/2)^3 + 5h(1/2)^4 + (1/2)^5 - (1/2)^5)/h

= h^4 + 5h^3(1/2) + 10h^2(1/2)^2 + 10h(1/2)^3 + 5(1/2)^4

Now as h —> 0 all of the terms with h in them go away and:

(((.5 + h)^5) - ((.5)^5))/h —> 5(1/2)^4 = 5/16.

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