find derivitive of 1/square root of x
1/square root of x = x^(-1/2)
The derivative is -1/2 x^(-3/2)
can you keep going on the problem to not have negative exponents?
You can also use that the derivative of
1/f(x) is given by:
-f'(x)/[f(x)]^2
If you take f(x) = sqrt(x) and use that f'(x) = 1/[2*sqrt(x)], you find that the derivative is given by:
-1/2 * 1/[sqrt(x)]^3
To simplify further and avoid negative exponents, we can rewrite the expression as follows:
-1/2 * 1/[sqrt(x)]^3
= -1/2 * 1/[(sqrt(x))^3]
= -1/2 * 1/(x^(1/2))^3
= -1/2 * 1/x^(3/2)
= -1/(2x^(3/2))
Therefore, the derivative of 1/sqrt(x) is -1/(2x^(3/2)).