I will use the quotient rule here.
d/dx(f(x)/g(x) = g(x)*f'(x) - f(x)*g'(x)/[g(x)]2
x3*1/x - ln(x)*3x2/(x3)2
x3/x - 3ln(x)x2/x6
x2 - 3ln(x)x2/x6
= - 3ln(x)/x4
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The derivative of e^u(x) with respect to x: [du/dx]*[e^u(x)]For a general exponential: b^x, can be rewritten as b^x = e^(x*ln(b))So derivative of b^x = derivative of e^u(x), where u(x) = x*ln(b).Derivative of x*ln(b) = ln(b). {remember b is just a constant, so ln(b) is a constant}So derivative of b^x = ln(b)*e^(x*ln(b))= ln(b) * b^x(from above)
if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.
I don't believe that the answer is ln(x)x^(ln(x)-2), since the power rule doesn't apply when you have the variable in the exponent. Do the following instead:y x^ln(x)Taking the natural log of both sides:ln(y)ln(x) * ln(x)ln(y) ln(x)^2Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * yFinally, substitute in the first equation, y x^ln(x):y' 2 ln(x) / x * x^ln(x)y'2 ln(x) * x ^ (ln(x) - 1)Sorry if everything is formatted really badly, this is my first post on answers.com.
x^(ln(2)/ln(x)-1)
1/x2