Derivative of xcosx: Proof by First Principle, Product Rule

The derivative of xcos x is equal to cosx – xsinx. The function xcos x is the product of x with its cosine. In this article, we will learn how to find the differentiation of xcos x using the following methods:

• Product rule of derivatives
• First principle of derivatives.

Derivative of xcosx Formula

The derivative of xcosx formula is given by d(xcosx)/dx = cosx – xsinx, where the differentiation is considered with respect to x. In the next two sections, we will prove this formula using the product rule of derivatives and the first principle of derivatives, that is, the definition of limits.

Derivative of xcosx by Product Rule

If f(x) is a product of two functions g(x) and h(x), that is, f(x) = g(x)h(x), then the derivative of f(x) by the product rule of derivatives is as follows.

$f'(x)$ $=g(x) h'(x) + h(x) g'(x)$,

where $’$ denotes the derivative with respect to x.

Put g(x) = x and h(x) = cosx ⇒ g$’$(x) = 1 and h$’$(x) = -sinx.

So the derivative of f(x)=xcos x is equal to

$(x\cos x)’$

= $x (\cos x)’ + \cos x (x)’$

= $-x \sin x + \cos x \cdot 1$

= cos x – x sin x.

Hence, the derivative of x cos x is cos x – x sin x obtained using the product rule of derivatives.

Also Read:

Derivative of sin3x: The derivative of sin3x is 3cos3x.

Derivative of log2x: The derivative of log2x is 1/x.

Derivative of 1: The derivative of 1 is zero.

Derivative of 1/x²: The derivative of 1/x² is -2/x³.

Derivative of 1/x³: The derivative of 1/x³ is -3/x⁴.

Derivative of xcosx by First Principle

We know that the derivative of a function f(x) by the first principle, that is, by the limit definition is given as follows.

$f'(x)=\lim\limits_{h\to 0}\dfrac{f(x+h)-f(x)}{h}$.

Put f(x) = x cos x.

So the derivative of xcosx from first principle is equal to

(xcos x)$’$ $=\lim\limits_{h\to 0}\dfrac{(x+h)\cos(x+h)-x\cos x}{h}$

= $\lim\limits_{h \to 0} \dfrac{x\cos(x+h)+h\cos(x+h)-x\cos x}{h}$

Rearrange the numerator and apply the sum rule of limits. Then the above limit

= $\lim\limits_{h \to 0} \dfrac{x\cos(x+h)-x\cos x}{h}$ $+\lim\limits_{h \to 0} \dfrac{h\cos(x+h)}{h}$

Applying the formula cos a – cos b = 2 sin(a+b)/2 sin(b-a)/2, the above limit is equal to

$\lim\limits_{h \to 0} \dfrac{x[2\sin \frac{x+h+x}{2} \sin \frac{x-x-h}{2}]}{h}$ $+\lim\limits_{h \to 0} [\cos(x+h)]$

= $-\lim\limits_{h \to 0} \dfrac{2x\sin (x+\frac{h}{2}) \sin \frac{h}{2}}{h}$ $+\cos(x+0)$ as we know that sin(-x)=-sin x.

= $-\lim\limits_{h \to 0} \Big[x\sin (x+\frac{h}{2}) \dfrac{\sin h/2}{h/2} \Big]$ $+\cos x$

[Let t=h/2. Then t→0 as h→0]

= $-x\sin (x+\frac{0}{2})\lim\limits_{t \to 0} \dfrac{\sin t}{t}$ $+\cos x$

= $-x\sin (x+0) \cdot 1+\cos x$ as the limit of sin(x)/x is 1 when x→0.

= -xsin x +cos x.

Thus, the derivative of xcos x is equal to (cosx-xsinx) and this is obtained by the first principle of differentiation.

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