Could someone help me with an explanation?
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I can help!
think of x/sqrt(x) as x^(2/2) / x^(1/2)
simplifying this gives x^(1/2)
much simpler derivative now.
d/dx x^1/2
=1/(2*x^1/2)
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hope this helps!
think of x/sqrt(x) as x^(2/2) / x^(1/2)
simplifying this gives x^(1/2)
much simpler derivative now.
d/dx x^1/2
=1/(2*x^1/2)
***
hope this helps!
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f(x) = x / [x^(1/2)]
f(x) = x[x^(1/2)] / x (rationalize denominator)
f(x) = (x/x)[x^(1/2)]
f(x) = x^(1/2)
Derivative:
f'(x) = (1/2)[x^(-1/2)]
f'(x) = (1/2){ [x^(1/2)] / x } (rationalize denominator)
f'(x) = sqrt(x) / 2x
The square root is the same as raising it to the half power.
f(x) = x[x^(1/2)] / x (rationalize denominator)
f(x) = (x/x)[x^(1/2)]
f(x) = x^(1/2)
Derivative:
f'(x) = (1/2)[x^(-1/2)]
f'(x) = (1/2){ [x^(1/2)] / x } (rationalize denominator)
f'(x) = sqrt(x) / 2x
The square root is the same as raising it to the half power.
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You could use the quotient rule but it will be more work than needed.
y = (x)^1 * (x)^ -1/2 = x^1/2 Multiplied both exponents and got x^1/2 .
y = x^1/2
dy/dx = (1/2) x^-1/2
y = (x)^1 * (x)^ -1/2 = x^1/2 Multiplied both exponents and got x^1/2 .
y = x^1/2
dy/dx = (1/2) x^-1/2
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y = x^(1/2)---> y ' = ...