How can I differentiate from the first principles cos x^2?

Answer 1

You want to calculate (f(x+h)-f(x))/h and then take the limit as h→0. So f(x+h)=cos(x+h)=cos(h)cos(x)-sin(h)sin(x).

so f(x+h-f(x)= cos(x+h)=cos(h)cos(x)-sin(h)sin(x)-cos(x). Now factor out cos(x), cos(x)(cos(h)-1)-sin(h)sin(x)

This is the numerator so we have to divide this by h

(cos(x)(cos(h)-1)-sin(h)sin(x))/h now when you let h go to 0 whats left is -sin(h)sin(x)/h whose limit is -sin(x)

lim h→0 of sin(h)/h=1

Let me know if you have questions

Jim

Answer 2

f(x) = cos(x2)

lim (as h ⇒ 0) = ((cos(x+h)²) - cos(x²)) / h

= ((cos(x² + 2xh + h²)) - cos(x²))/h use: cos(x+y) = cosxcosy - sinxsiny

= (cos(x²)cos(2xh + h²) - sin(x²)sin(2xh + h²) - cos(x²)) / h

= (cos(x²) x (cos(2xh + h²) - 1) /h) - (sin(x²)sin(2xh + h²) /h) multiply both terms by (2x+h / 2x+h)

=(cos(x²) x (cos(2xh + h²) - 1)(2x+h) /h(2x+h)) - (sin(x²)sin(2xh + h²)(2x+h) /h(2x+h))

use: lim t ⇒0 (cost -1) /t = 0 use: lim t ⇒0 sint /t =1

Therefore: lim h⇒0 (cos(2xh + h²) - 1)(2x+h) /h(2x+h)) = 0

lim h⇒0 (sin(x²)sin(2xh + h²)(2x+h) /h(2x+h)) = 1

lim h ⇒0 2x+h = 2x

cos(x²)(0)(2x) - sin(x²)(1)(2x) = -2xsin(x²)

Answer 3

d/du (cos(u)) = -sin(u)*du

In your case.....d/dx(cos(x²)) = -sin(x²)*2x