Recent content by TommG

I made some corrections ln y = ln \sqrt{x(x+1)} ln y = \frac{1}{2}ln x(x+1) ln y = \frac{1}{2}ln x + ln(x+1) \frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1} \frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1} \frac{1}{y}= \frac{1}{x} + \frac{1}{x+1} \frac{dy}{dx}= y(\frac{1}{x}...

Need to find derivative using logarithmic differentiation y = \sqrt{x(x+1)} My attempt ln y = ln \sqrt{x(x+1)} ln y = \frac{1}{2}ln x(x+1) ln y = \frac{1}{2}ln x + ln(x+1) \frac{1}{y}= (\frac{1}{2}) \frac{1}{x} + \frac{1}{x+1} \frac{1}{y}= \frac{1}{2x} + \frac{1}{x+1} \frac{dy}{dx}=...

Yeah it is correct

Need to find derivative y = θ(sin(ln θ)) + cos(ln θ) My work θ(cos(ln θ))(1/θ) + sin(ln θ) + (-sin(ln θ)(1/θ)) (θcos(ln θ))/θ] + sin(ln θ) + ( (- sin(ln θ))/θ) cos(ln θ) + [θsin(ln θ) - sin(ln θ)]/ θ answer in book is 2cos(lnθ)

thank you for your help I have figured it out

Need to find derivative r = sin(θ2)cos(2θ) answer in book -2sin(θ2)sin(2θ)+2θcos(2θ)cos(θ2)My attempt tried using the product rule (sin(θ2))(-sin(2θ)) + cos(2θ)cos(θ2) I got stuck right here

I am not sure if I did it right but I think I got the answer. x2sin4(x) 2xsin4 took derivative of just x2 4x2sin3x derivative of just power of sin cosx derivative of just sin 2xsin4(x)+4x2sin3(x)cos(x) xcos-2(x) cos-2 derivative of x -2xcos-3(x) derivative of power of cos...

Need to find the Derivative using the chain rule y = x2sin4(x) + xcos-2(x) I am not sure where to start. answer in book is 2xsin4(x) + 4x2sin3(x)cos(x) + cos-2(x) +2xcos-3(x) xsin(x)

I have to do it the hard because my teacher said I have to when he gives us a quiz. I put everything over a common denominator because of the rule in the book. Here is a pic I took from my book. https://www.dropbox.com/s/ca3l50hmgh56n41/IMG_20140606_121016.jpg

I have to do it the hard way. I have to follow the derivative quotient rule. Still need help if anyone could help out.

Well I figured out this problem but need help with another. Thanks ray for the advice. I could use help with a new problem. Have to find the derivative. v = \frac {1 + x - 4\sqrt{x}} x My attempt v = \frac {1 + x - 4\sqrt{x}} x v = \frac {x(1-\frac{2}{√x}) - (1+x -4...

yes I do. I deleted my attempt because I don't think I can write it well enough so people can understand it. Is it ok if I post a pic of it?

Need to find derivative f(s) = [(√s) -1]/[(√s) + 1] answer in book is f'(s) = 1/[√s(√s+1)^2]

I am sorry it is the first one p = \frac{1}{\sqrt{q+1}}

find the indicated derivative dp/dq if p = 1/(√q+1) I apologize ahead of time if you can't read my work. my work [(1/(√(q+h+1))) - (1/(√(q+1))] \divh [((√(q+1)) - (√(q+h+1)))/((√(q+h+1))(√(q+1)))] \divh [(q+1-q-h-1)/(((√q+h+1)(√q+1))((√q+1)+(√q+h+1)))]\div h...