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What is the derivative of lnx raised to lnx?
start by setting y=lnx^lnx take ln of both sides lny=lnx(ln(lnx)) differentiate dy/dx(1/y)=(1+ln(lnx))/x dy/dx=y(1+ln(lnx))/x we know that y=lnx^lnx so we can just substatute back in dy/dx=(lnx^lnx)*(1+ln(lnx))/x
What is the Second derivative of lnx?
-1/x2
How do you solve for x when the natural log of the square root of x equals the square root of the natural log of x?
ln(√x)=√(lnx) because √x = x^(1/2), ln(x^(1/2))=√(lnx) using a logarithmic property, we can say that .5(lnx)=√(lnx) now, pretend that lnx=y .5y=√y square both sides .25y^2=y subtract y from both sides .25y^2 -y=0 factor y(.25y - 1)=0 so either y=0 or .25y -1 =0 If .25y -1=0, then y=4 so lnx=0 or lnx=4 lnx cannot equal zero because lnx=0 means e^x=0 and that is impossible. Now, we are left with lnx=4 Isolate x by making both sides of the equation powers of e: e^(lnx)=e^4 x=e^4, which is approximately 54.6 Lastly, check this answer by plugging e^4 back into the original equation: ln(√(e^4))=√(ln(e^4)) ln(e^2)=√(4(lne)) 2lne=2√1 2(1)=2 2=2 There you go!
What is the derivative of lnx?
Derivative of lnx= (1/x)*(derivative of x) example: Find derivative of ln2x d(ln2x)/dx = (1/2x)*d(2x)/dx = (1/2x)*2===>1/x When the problem is like ln2x^2 or ln-square root of x...., the answer won't come out in form of 1/x.
How do you evaluate definite integrals?
In order to evaluate a definite integral first find the indefinite integral. Then subtract the integral evaluated at the bottom number (usually the left endpoint) from the integral evaluated at the top number (usually the right endpoint). For example, if I wanted the integral of x from 1 to 2 (written with 1 on the bottom and 2 on the top) I would first evaluate the integral: the integral of x is (x^2)/2 Then I would subtract the integral evaluated at 1 from the integral evaluated at 2: (2^2)/2-(1^2)/2 = 2-1/2 =3/2.
