0
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)
What else can I help you with?
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 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 x to the power of ln x?
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.
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)
What is second order differential equation?
The highest order of derivative is 2. There will be a second derivative {f''(x) or d2y/dx} in the equation.
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 4th derivative of lnx?
The fourth derivative of ( \ln(x) ) can be determined by first calculating its derivatives. The first derivative is ( \frac{1}{x} ), the second derivative is ( -\frac{1}{x^2} ), the third derivative is ( \frac{2}{x^3} ), and the fourth derivative is ( -\frac{6}{x^4} ). Thus, the fourth derivative of ( \ln(x) ) is ( -\frac{6}{x^4} ).
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
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 Ln10?
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.
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 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 ln x?
The anti-derivative of ( \ln x ) can be found using integration by parts. Let ( u = \ln x ) and ( dv = dx ), then ( du = \frac{1}{x} dx ) and ( v = x ). Applying integration by parts, we get: [ \int \ln x , dx = x \ln x - \int x \cdot \frac{1}{x} , dx = x \ln x - x + C, ] where ( C ) is the constant of integration. Thus, the anti-derivative of ( \ln x ) is ( x \ln x - x + C ).
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.
