e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
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)]
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
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.
the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x
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.
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/x
if f(x)=kx, f'(x)=ln(k)*kx. Therefore, the derivative of 2x is ln(2)*2x.
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
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)]
e^[ln(x^2)]=x^2, so your question is really, "What is the derivative of x^2," to which the answer is 2x.
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
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.
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)