Also I would like to know how to set this problem up in finding the answer: Find the second derivative. y= sqrt(3x+4) Thanks, P.S. I tried taking 3x+4 and squaring it to the 1/2. I thought that might've been the same, then I took
How would I go about finding the derivative of f(x)=ln(x)(7x-2)^3 first=ln(x) second= (7x-2)^3 Would I do first times the derivative of the second + second times the derivative of the first? If so, what is the derivative of lnx?
Okay, I want to find the derivative of (x^x)^(x^x)... Well, I already went through the process of finding the derivative of just x^x (I used ln to do this). This is what I found... dy/dx = x^x(lnx+1) So now I want to find the
I need help with finding first and second derivative--simplify your answer. Y=xtan(x) I solved first derivative xsec^2(x) + tan(x) I don't understand how to get the first...can someone show me the steps with explanation?
For positive a, b, the potential energy, U, of a particle is given by U = b (a^2 over x^2 - a over x) for x is greater than 0 a) find the x-intercept b) find the asymptotes c)compute the local minimum d) sketch the graph I don't
Okay, how would you go about finding the area of a curve from 1 to 4, when y=2x+(2/(x^2))?? It's not like the problem I asked before because here, you cannot use substitution. I tried using 2x for u and x^2 for du but it won't
y=ln(x+1) +ln(x-1) find dy/dx I have trouble with this problem since i'm not that familar with finding the derivative when they involve ln. Please show me how to solve this problem. I should have gotten 2x/(x^2-1) The derivative