Recent content by LizzieL

Thank you for your help! I think I'm getting there: \frac{dy}{dx}= \frac {2.13}{2\sqrt{x}} \Rightarrow dx = \frac {2\sqrt{x}}{2.13}dy \Rightarrow \int sin(y) \cdot \frac {2}{2.13}\Big(\frac {y-2.4}{2.13}\Big)dy = \frac {-2.4}{2.13^2} cos(y) + \frac {2}{2.13^2}\int y sin(y) dy...

Ok, thanks! So, is this right or close to right?: y^,= 2.13x^{1/2}= 2.13(\frac{1}{2})x^{-1/2}=\frac{2.13}{2\sqrt{x}} And proceeding, dy= \frac{2.13}{2\sqrt{x}} dx ? How do I move from here?

I'm stuck! :cry: I'm desperately trying to figure out the derivative of y=2.13\sqrt{x}+2.4 but I can't seem to get it right. I'm starting off like this: y'=(2.13\sqrt{x}+2.4)'= 2.13x^{1/2} ? How do I calculate this?

Normally I don't think I have a problem understanding the chain rule, but it's all backwards when in integration context, and it confuses me. My textbook is not covering this topic properly (in my opinion). So to answer your question, I do not know how to apply the chain rule here... :shy:

Thank you! Ok, so I have x=\big( \frac {y-2.4}{2.13}\big)^2 And, dx= 2 \big( \frac {y-2.4}{2.13}\big) dy Is this what you mean? I don't think I understood the last part you mentioned though.

Homework Statement I have the integral \int sin(2.13\sqrt{x}+2.4)\,dx I'm supposed to use the substitution y=2.13\sqrt{x}+2.4, aka. \sqrt{x}=\frac {y-2.4}{2.13} to gain the following description of the integral: \int sin(2.13\sqrt{x}+2.4)\,dx = E cos(y) + F\int y sin(y)\,dy I have...

Homework Statement I have L = \lim_{x\rightarrow 0} \Big( {\frac{\cos(1.92x)-1} {e^{2.33x} - 1 -2.33x}} \Big) I'm meant to use L'Hôpital's rule finding the limit, maybe twice. The Attempt at a Solution So, there's clearly something I have misunderstood, and hoping you might tell...

I got it! Thanks :approve:

Homework Statement Function f(x) = KxL K= 1.78 L= -1.39 Problem 1: Find f'(x). ____________________ v= 0.89 w= 0.5 "v" and "w" are two points located on the x-axis. Problem 2: Calculate f'(v). ____________________ Problem 3: Find the equation of the tangent line of f(x) over the point "w". The...

Oh my, you guys really are angels sent from up above. Thanks a whole lot! Also for the welcome :smile: So, I have a similar problem: log(y2) = 1.85t + 1.56 = log(e)μ(t-a) I cannot use the same principle as before, right? Or can I? If I try again... 1.85 = log(e)μ 1.56 = -log(e)μa...

Oh, of course the logarithm of y is given! Why didn't I see that before. Thanks for your reply. The base of the logarithm is 10, forgot to say. But, how do I solve it from here? This is my first go at a problem like this, so any help is appreciated. How can I solve this when I have no values...

Homework Statement Ok, so I have an unknown exponential function: y1 = f1(t) By measuring the values of t and yi, a linear connection is generated between Yi (=log (yi)) and t: Yi = Ait + Bi. A1 = -2.12 B1 = 1.96 Problem 1: Describe f1 in the following matter: y1 = f1(t) = Ceλt...