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Mathematics 16 Online
OpenStudy (anonymous):

prove : d/dx(square root of u) is 1/2squareroot of u (du/dx)

hartnn (hartnn):

isn't this direct application of chain rule?

OpenStudy (barrycarter):

Hint: definition of derivative

sam (.sam.):

yeah you have to use the definition of derivative

OpenStudy (anonymous):

how?

OpenStudy (anonymous):

d (u)^1/2 dx = (d (u)^1/2 du) (du/dx)

OpenStudy (anonymous):

d (u)^1/2 dx = (d (u)^1/2/ du) (du/dx)

OpenStudy (anonymous):

use d (x^n)/dx = n x^(n-1)

OpenStudy (anonymous):

then?

OpenStudy (anonymous):

d (u)^1/2/ du = (1/2)(u)^(1/2 - 1)

OpenStudy (anonymous):

and put the value of d (u)^1/2/ du in this d (u)^1/2 dx = (d (u)^1/2/ du) (du/dx)

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