Question

Find the derivate of x.cosx by
"first principle"..


class 11 math..

please help __/¡\__
​

Answer 1

$\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}$

Here the concept of First Principal of Differentiation has been used. We know that the rate of change of y with respect to x is the first principal of differentiation. First we will apply values in formula, then simply it. Finally we will convert it into simplest form to find the answer.

Let's do it  !!

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★ Formula Used :-

$\\\;\;\boxed{\sf{\dfrac{dy}{dx}\;=\;\bf{\lim_{h\:\to\:0}\;\;\dfrac{f(x\;+\;h)\;-\;f(x)}{h}}}}$

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★ Solution :-

→ f(x) = x.cos x

Using First Principal of Differentiation, we get,

$\\\;\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{dy}{dx}\;=\;\bf{\lim_{h\:\to\:0}\;\;\dfrac{f(x\;+\;h)\;-\;f(x)}{h}}}}$

$\\\;\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\:f(x)\;=\;\bf{\lim_{h\:\to\:0}\;\;\dfrac{f(x\;+\;h)\;-\;f(x)}{h}}}}$

$\\\;\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\:f(x)\;=\;\bf{\lim_{h\:\to\:0}\;\;\dfrac{(x\;+\;h)\cos(x\;+\;h)\;-\;(x\cos\:x)}{h}}}}$

$\\\;\;\displaystyle{\sf{:\rightarrow\;\;\dfrac{d}{dx}\:f(x)\;=\;\bf{\lim_{h\:\to\:0}\;\;\dfrac{(x\;+\;h)\:(\cos\:x\;\cos\:h\;-\;\sin\:x\;\sin\:h)\;-\;(x\cos\:x)}{h}}}}$

$\\\displaystyle{\sf{\odot\;\dfrac{d}{dx}f(x)\;=\;\bf{\lim_{h\to 0}\dfrac{(x\cos x\cos h\;-\;x\sin x\sin h)\;+\;(h\cos x\cos h\;-\;h\sin x\sin h)\:-\:(x\cos\:x)}{h}}}}$

$\\\displaystyle{\sf{\odot\;\dfrac{d}{dx}f(x)\;=\;\bf{\lim_{h\:\to\:0}\;\;\dfrac{x\cos x\cos h\;-\;(x\cos x)\;-\;x \sin x\sin h\;+\;h\:\cos\:x\;\cos\:h\;-\;h\:\sin\:x\;\sin\:h}{h}}}}$

$\\\displaystyle{\sf{\odot\;\dfrac{d}{dx}f(x)\;=\;\bf{x\cos x\lim_{h\to 0}\dfrac{\cos h\;-\;1}{h}\;-\;x\sin x\lim_{h\to 0}\dfrac{\sin h}{h}+\cos x\lim_{h\to 0}\cos h+\sin x\lim_{h\to 0}\sin h}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;\dfrac{d}{dx}\:f(x)\;=\;\bf{x\cos\:x\;\lim_{h\:\to\:0}\;\dfrac{-2\sin^{2}\:\bigg(\dfrac{h}{2}\bigg)}{\bigg(\dfrac{h^{2}}{4}\bigg)}\;\times\;\dfrac{h}{4}\;-\;x\sin\:x(1)\;+\;\cos\:x(1)\;+\;\sin\:x(0)}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;\dfrac{d}{dx}\:f(x)\;=\;\bf{x\cos\:x\;\lim_{h\:\to\:0}\;\dfrac{-\:h}{2}\;-\;x\sin\:x(1)\;+\;\cos\:x(1)\;+\;\sin\:x(0)}}}$

$\\\;\;\displaystyle{\sf{:\Longrightarrow\;\;\dfrac{d}{dx}\:f(x)\;=\;\bf{x\cos\:x\;(0)\;-\;x\sin\:(x)\;+\;\cos\:(x)}}}$

$\\\;\;\displaystyle{\bf{:\Longrightarrow\;\;\dfrac{d}{dx}\:f(x)\;=\;\bf{-\;x\sin\:(x)\;+\;\cos\:(x)}}}$

$\\\;\underline{\boxed{\tt{Hence,\;\;derivative\;\;of\;\;x.\:\cos\:x\;=\;\bf{-\:x\sin\;(x)\;+\;\cos\:(x)}}}}$

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★ More to know :-

• Second Principal of Differentiation :

If f(x) is the given function, then the second derivative is f'(x).

$\\\;\sf{\leadsto\;\;\dfrac{d}{dx}\:f(x)\;=\;\dfrac{d^{2}y}{dx^{2}}}$

• Third Principal of Differentiation :

If f(x) is the given function, then the third derivative is f"(x).

$\\\;\sf{\leadsto\;\;\dfrac{d}{dx}\:f(x)\;=\;\dfrac{d^{3}y}{dx^{3}}}$