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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.
