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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)^2

Take the derivative of both sides, using the chain rule:1/y * y' 2 ln(x) / xy' 2 ln(x)/ x * y

Finally, 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.

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Q: What is the derivative of x to the power of ln x?
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What is the derivative of 2 to the power of x?

if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.


How do you differentiate exponential function?

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)


What is 2 divided by x as a power of x?

x^(ln(2)/ln(x)-1)


What is the second derivative of x ln x?

f(x)=xln(x) this function is treated as u*v u=x v=ln(x) The derivative of a product is f'(x)=u*v'+v*u' plugging the values back in you get: f'(x)=(x*dlnx/x)+(ln*dx/dx) The derivative of lnx=1/x x=u dlnu/dx=(1/u)*(du/dx) dx/dx=1 x=u dun/dx=nun-1 dx1/dx=1*x1-1 = x0=1 f'(x)=x*(1/x)+lnx*1 f'(x)=1+lnx Now for the second derivative f''(x)=d1/dx+dlnx/dx the derivative of a constant, such as 1, is 0 and knowing that the derivative of lnx=1/x you get f''(x)=(1/x)


Differentiate log x?

The derivative of ln x, the natural logarithm, is 1/x.Otherwise, given the identity logbx = log(x)/log(b), we know that the derivative of logbx = 1/(x*log b).ProofThe derivative of ln x follows quickly once we know that the derivative of ex is itself. Let y = ln x (we're interested in knowing dy/dx)Then ey = xDifferentiate both sides to get ey dy/dx = 1Substitute ey = x to get x dy/dx = 1, or dy/dx = 1/x.Differentiation of log (base 10) xlog (base 10) x= log (base e) x * log (base 10) ed/dx [ log (base 10) x ]= d/dx [ log (base e) x * log (base 10) e ]= [log(base 10) e] / x= 1 / x ln(10)

Related questions

What is the derivative of e the the power ln x?

y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1


What is the derivative of ln x to the power of 2?

If the function is (ln x)2, then the chain rules gives us the derivative 2ln(x)/x, with the x in the denominator. If the function is ln (x2), then the chain rule gives us the derivative 2/x.


What is the derivative of e to the power ln x squared?

e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.


What is the derivative of 2 to the power of x?

if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.


What is the derivative of lnlnx?

1/xlnx Use the chain rule: ln(ln(x)) The derivative of the outside is1/ln(x) times the derivative of the inside. 1/[x*ln(x)]


What is the derivative of x to the power of e?

e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.


How do you differentiate exponential function?

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)


How do you do exponential functions?

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)


What is the Derivative of 500 ln x plus 1?

the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x


What is the derivative of X-1 to the power of X?

d/dx (X - 1)x = (X - 1)x ln(X - 1) * x = X(X- 1)x ln(X - 1) -------------------------


What is the derivative of 2lnx?

The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x


What is the anti-derivative of 5 to the x power?

ex and ln(x) are inverse functions. With this you can get 5x = eln(5^x) Therefore you can anti-differentiate this to get eln(5^x)/(ln(5x)) Which equals 5x/ln(5x)