# Deivative by first principle.

• Sep 12th 2008, 11:56 AM
tariq_h_tauheed
Deivative by first principle.
I am not able to find the derivative of the following function by first principle. I need the entire procedure of the first principle. Thanx in advance.

f(x)= x^2.cosx
• Sep 12th 2008, 12:03 PM
arbolis
$\displaystyle (u\times v)'(x)=u'(x)v(x)+u(x)v'(x)$.
Put $\displaystyle u(x)=x^2$ so $\displaystyle u'(x)=2x$.
Put $\displaystyle v(x)=\cos (x)$ so $\displaystyle v'(x)=-\sin (x)$. I let you finish this, it's just 2 multiplications and an addition.
EDIT: Oops... sorry I didn't know what what the "first principle".
• Sep 12th 2008, 12:08 PM
guangzhao
You can differentiate using the chain rule.

Use the identity e^(x^2)=2e^x

so e^y=2e^(x.cosx)

Differentiating,
e^y(dy/dx)=2(cosx+(-xsinx))e^(x.cosx)

Then just move e^y over and sub it as e^(x^2.cosx)
• Sep 12th 2008, 12:08 PM
topsquark
Quote:

Originally Posted by tariq_h_tauheed
I am not able to find the derivative of the following function by first principle. I need some urgent help. Thanx in advance.

f(x)= x^2.cosx

Guys, the question is from first principles, so we don't have access to the chain rule.

$\displaystyle f'(x) = \lim_{h \to 0} \frac{(x + h)^2~cos(x + h) - x^2~cos(x)}{h}$

Recall that $\displaystyle cos(x + h) = cos(x)~cos(h) - sin(x)~sin(h)$, so expanding we get
$\displaystyle = \lim_{h \to 0} \frac{(x^2 + 2hx + h^2)~(cos(x)~cos(h) - sin(x)~sin(h)) - x^2~cos(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{(x^2 + 2hx + h^2)~cos(x)~cos(h) - (x^2 + 2hx + h^2) sin(x)~sin(h) - x^2~cos(x)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{(2hx + h^2)~cos(x)~cos(h) - (x^2 + 2hx + h^2) sin(x)~sin(h)}{h}$

$\displaystyle = \lim_{h \to 0} \frac{(2hx + h^2)~cos(x)~cos(h)}{h} - \lim_{h \to 0}\frac{(x^2 + 2hx + h^2) sin(x)~sin(h)}{h}$

See if you can finish this.

-Dan