0
By parts. You'll get
((x^2)/2) * (lnx - 1/2) + c
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∙ 12y ago
Anonymous
x4 lnxdx
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What is the second derivative of x ln x?
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)
Integrate of ln x squared?
int(ln(x2)dx)=xln|x2|-2x int(ln2(x)dx)=x[(ln|x|-2)ln|x|+2]
How do you integrate cosine squared times sine?
Trying to integrate: cos2x sin x dx Substitute y = cos x Then dy = -sin x dx So the integral becomes: -y2dy Integrating gives -1/3 y3 Substituting back: -1/3 cos3x
What is the integral of tan squared x?
Note that for sec²(x) - tan²(x) = 1, we have: -tan²(x) = 1 - sec²(x) tan²(x) = sec²(x) - 1 Rewrite the expression as: ∫ (sec²(x) - 1) dx = ∫ sec²(x) dx - ∫ 1 dx Finally, integrate each expression to get: tan(x) - x + K where K is the arbitrary constant
How do you integrate of e -2x?
To integrate e^(-2x)dx, you need to take a u substitution. u=-2x du=-2dx Since the original integral does not have a -2 in it, you need to divide to get the dx alone. -(1/2)du=dx Since the integral of e^x is still e^x, you get: y = -(1/2)e^(-2x) Well, that was one method. I usually solve easier functions like this by thinking how the function looked like before it was differentiated. I let f(x) stand for the given function and F(x) stand for the primitive function; the function we had before differentiation (the integrated function). f(x)= e-2x <-- our given function F(x)= e-2x/-2 <-- our integrated function Evidence: F'(x)= -2e-2x/-2 = e-2x = f(x) Q.E.D It's as simple as that.
What is the derivative of lnx raised to 4?
ln(x4)?d/dx(ln(u))=1/u*d/dx(u)d/dx(ln(x4))=[1/x4]*d/dx(x4)-The derivative of x4 is:d/dx(x4)=4x4-1d/dx(x4)=4x3d/dx(ln(x4))=[1/x4]*(4x3)d/dx(ln(x4))=4x3/x4d/dx(ln(x4))=4/x(lnx)4?Chain rule: d/dx(ux)=x(u)x-1*d/dx(u)d/dx(lnx)4=4(lnx)4-1*d/dx(lnx)d/dx(lnx)4=4(lnx)3*d/dx(lnx)-The derivative of lnx is:d/dx(ln(u))=1/u*d/dx(u)d/dx(lnx)=1/x*d/dx(x)d/dx(lnx)=1/x*(1)d/dx(lnx)=1/xd/dx(lnx)4=4(lnx)3*(1/x)d/dx(lnx)4=4(lnx)3/x
What is the derivative of x5lnx?
x5lnx?d/dx (uv)=u*dv/dx+v*du/dxd/dx (x5lnx)=x5*[d/dx(lnx)]+lnx*[d/dx(x5)]-The derivative of lnx is:d/dx(lnu)=(1/u)*[d/dx(u)]d/dx(lnx)=(1/x)*[d/dx(x)]d/dx(lnx)=(1/x)*[1]d/dx(lnx)=(1/x)-The derivative of x5 is:d/dx (xn)=nxn-1d/dx (x5)=5x5-1d/dx (x5)=5x4d/dx (x5lnx)=x5*[1/x]+lnx*[5x4]d/dx (x5lnx)=[x5/x]+5x4lnxd/dx (x5lnx)=x4+5x4lnx
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 derivative and second derivative of lnx to the third power?
If you mean: y =(lnx)3 then: dy/dx = [3(lnx)2]/x ddy/dx = [(6lnx / x) - 3(lnx)2] / x2 If you mean: y = ln(x3) Then: dy/dx = 3x2/x3 = 3/x = 3x-1 ddy/dx = -3x-2 = -3/x2
What is the derivative of lnx squared?
I do not see why the chain rule would not work here. d/dx (inx)^2 = 2(lnx) * 1/x = 2(lnx)/x
What is the derivative of natural log?
d/dx lnx=1/x
What is the second derivative of x ln x?
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 is the derivative of cos x raised to the x?
cos(xx)?d/dx(cosu)=-sin(u)*d/dx(u)d/dx(cos(xx))=-sin(xx)*d/dx(xx)-The derivative of xx is:y=xx ;You have to use implicit derivation because there is no formula for taking the derivative of uu.lny=lnxxlny=xlnx-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)-d/dx(uv)= u*dv/dx+v*du/dxTherefore:(1/y)*dy/dx=x*[(1/x)*d/dx(x)]+lnx(d/dx(x))-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1(1/y)*dy/dx=x*[(1/x)*(1)]+lnx(1)(1/y)*dy/dx=x*[(1/x)]+lnx(1/y)*dy/dx=(x/x)+lnx(1/y)*dy/dx=1+lnxdy/dx=y(1+lnx) ;Multiply y to both sidesdy/dx=xx(1+lnx) ;y=xx, so replace the y with xxd/dx(cos(xx))=-sin(xx)*[xx*(1+lnx)]d/dx(cos(xx))=-(1+lnx)*xx*sin(xx)(cosx)x?Again with the implicit derivation:y=(cosx)xlny=x*ln(cosx)(1/y)*dy/dx=x[d/dx(lncosx)]+lncosx(d/dx(x))(1/y)*dy/dx=x[(1/cosx)*(-sinx)(1)]+lncosx(1) ;The derivative of lncosx is (1/cosx)*d/dx(cosx). The derivative of cosx is (-sinx)*d/dx (x). The derivative of x is 1.(1/y)*dy/dx=x[(1/cosx)*(-sinx)]+lncosx(1/y)*dy/dx=x[-tanx]+lncosx(1/y)*dy/dx=-xtanx+lncosxdy/dx=y(-xtanx+lncosx) ;Multiply both sides by ydy/dx=(cosx)x(-xtanx+lncosx) ;y=(cosx)x, replace all y's with (cosx)xdy/dx=(cosx)x(-xtanx+lncosx)=(cosx)x-1(cosx*lncosx-xsinx)
What is the Second derivative of 3.9625.lnx?
3.9625lnx?The first derivative is:d/dx(cu)=c*du/dx where c is a constant.d/dx(3.9625lnx)=3.9625*d/dx(lnx)-The derivative of lnx is:d/dx(lnu)=(1/u)*d/dx(u)d/dx(lnx)=(1/x)*d/dx(x)d/dx(3.9625lnx)=3.9625*[(1/x)*d/dx(x)]-The derivative of x is:d/dx(xn)=nxn-1d/dx(x)=1*x1-1d/dx(x)=1*x0d/dx(x)=1*(1)d/dx(x)=1d/dx(3.9625lnx)=3.9625*[(1/x)*1]d/dx(3.9625lnx)=3.9625*(1/x)d/dx(3.9625lnx)=3.9625/xThe second derivative of 3.9625lnx is the derivative of 3.9625/x=3.9625*x-1:d/dx(cu)=c*du/dx where c is a constant.d/dx(3.9625*x-1)=3.9625*d/dx(x-1)-The derivative of x-1 is:d/dx(xn)=nxn-1d/dx(x-1)=-1*x-1-1d/dx(x-1)=-1*x-2d/dx(x-1)=-1/x2d/dx(3.9625*x-1)=3.9625*(-1/x2)d/dx(3.9625*x-1)=-3.9625/x2
Integrate x 5x dx?
integrate(x5x dx) simplifies to integrate(5x^2 dx), and using the power rule of integration, add one to the power of x and divide the term by that number. Thus, x5x dx integrated is (5/3)x^3
What is the integral of x.lnx?
Do you want ∫x lnx dx? Let's call that I, which we now seek to find. The solution is I = ½ x2 lnx - ¼ x2; here is how we can find it: Let z = lnx. Then, x = ez, dx = ez dz, and dI = x lnx dx. Then, dI = (ez)(z)(ez dz) = ze2z dz = ¼ (2z e2z d(2z)). Thus, 4dI = wew dw, where we let w = 2z = 2 lnx. Now, d(wew) = ew dw + w d(ew) = ew dw + wew dw = ew dw + 4dI; hence, 4dI = d(wew) - ewdw = d(wew) - d(ew)­ = d((w - 1)ew). Therefore, 4I = (w - 1)ew = (2 lnx - 1)x2 = 2x2 lnx - x2; and I = ½ x2 lnx - ¼ x2, which is the answer we sought. Checking, we differentiate back, to confirm that I' = x lnx: I = ¼ x2(2 lnx - 1), whence, 4I' = (x2(2 lnx - 1))' = 2x(2 lnx - 1) + x2 (2/x) = 2x(2 lnx - 1) + 2x = 2x(2 lnx) = 4x lnx; thus, I' = x lnx, re-assuring us that we have integrated correctly.
What is the derivative of ln 1 plus x?
d/dx of lnx is 1/x Therefore the derivative is 1/(1+x)
