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)
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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.
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)
The highest order of derivative is 2. There will be a second derivative {f''(x) or d2y/dx} in the equation.