Recent content by rishi kesh

Thanks for that

This is the explanation i actually needed. But i want you to extend it a little bit, hopefully you will clear my doubt(please check my attachment). I will appreciate further reply from you :)

But why is it true for radians only? I solved on my calculator and this is what i saw : Tan (0.12) = 0.12 (where 0.12 is radians) But when x is in degrees Tan(0.12) = 0.00209 What is happening here?please explain.

Hi! In one of my textbook i saw the relation tan(x) = x where x is very small value and expressed in radians. I want to know why its true and how it actually works. I would appreciate someone's help :smile:

I have a picture below please have a look at it. The guy is using two conducting rods placed parallel to each other and placed below it are magnets. When battey is connected to rods and a conductor is placed over them it starts to move.That works according to flemmings left hand rule. I was...

Hey! I think i got it! When, f(x)= x^2 f(x+h)= (x+h)^2 If, f(x)= -x f(x+h)= -(x+h) = -x-h Is this right?

But how does that work? Why it isn't -x+h ?:oldconfused:

Homework Statement This is a silly question,but i have a problem.How do we solve derivative of -x using first principle of derivative. I know that if derivative of x w.r.t x is 1 then ofcourse that of -x should be -1. Also it can be solved by product rule taking derivative of -1.x .Homework...

Here is a picture of what I've solve. Please tell me if its correct.

Hi! I am sorry i tried but i couldn't solve it . I would greatly appreciate your help.

Hi! Please give me a little bit of intuition about this. I will solve the rest. Thanks in advance !

Homework Statement How do we find the derivative of function: y= √[(1-sinx)/(1+sinx)] This is the exercise problem from my textbook. I have not covered chain rule yet. So please you basic derivative rules to solve it. Homework Equations Here is the answer of derivative given in my textbook...

You explained it very clear now i got it!

Thank Oh yeah! But for me its cute :P

So when we have a negative feedback there is a very little difference between the voltages at inverting and non inverting inputs.and since op amps in practical have some finite gain the output will be amplified without rising up extremely high. Am i right?