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
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
The derivative of ln(10) is 1/10. This is because the derivative of the natural logarithm function ln(x) is 1/x. Therefore, when differentiating ln(10), the derivative is 1/10.
x (ln x + 1) + Constant
-1/ln(1-x) * 1/(1-X) or -1/((1-x)*ln(1-x))
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
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
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
Oh, dude, the third derivative of ln(x) is -2/(x^3). But like, who really needs to know that, right? I mean, unless you're planning on impressing your calculus teacher or something. Just remember, math is like a puzzle, except no one actually wants to put it together.
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.
(1/2)x = 2-x = exp (ln 2-x) = exp( -x ln 2). Since d/dx exp(x) = exp(x), we can use the chain rule to find that: d/dx (1/2)x = -(ln 2) exp(-x ln 2).
(1/2)x = 2-x = exp (ln 2-x) = exp( -x ln 2). Since d/dx exp(x) = exp(x), we can use the chain rule to find that: d/dx (1/2)x = -(ln 2) exp(-x ln 2).
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)]
Do you mean ln(x-2), or ln(x)-2? If it is ln(x-2): 1/(x-2) If it is ln(x)-2: 1/x
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)