Question

obtain derivatives of the following functions: 1) x sin x 2) x4 + cos x 3) x/sin
​

Answer 1

Answer:

The derivatives of given functions are as follows:

1) $\frac{d}{dx} (xsin(x)) = sin(x)+xcos(x)$

2) $\frac{d}{dx}(x^{4}+cos(x)) = 4x^{3}-sin(x)$

3) $\frac{d}{dx}(x/sin(x)) = \frac{sin(x) - xcos(x)}{sin^{2}(x) }$

Explanation:

1) $\frac{d}{dx} (xsin(x))$

We apply product rule here

$\frac{d}{dx} (xsin(x))$ = $(x)\frac{d}{dx} (sin(x))+(sin(x))\frac{d}{dx} (x)$

$\frac{d}{dx} (xsin(x)) = (x) (cos(x)) + sin(x)$

⇒ $\frac{d}{dx} (xsin(x)) = sin(x)+xcos(x)$

2) $\frac{d}{dx}(x^{4}+cos(x))$

We apply power rule here

$\frac{d}{dx}(x^{4}+cos(x)) = \frac{d}{dx} (x^{4})+\frac{d}{dx}(cos(x))$

$\frac{d}{dx}(x^{4}+cos(x)) = 4x^{3} + (-sin(x))\\\frac{d}{dx}(x^{4}+cos(x)) = 4x^{3} - sin(x)$

⇒ $\frac{d}{dx}(x^{4}+cos(x))=4x^{3} - sin(x)$

3) $\frac{d}{dx}(x/sin(x))$

We apply quotient rule here

$\frac{d}{dx}(x/sin(x)) = \frac{sin(x)\frac{d}{dx}(x)- x \frac{d}{dx}(sin(x))}{sin^{2}(x)}\\ \frac{d}{dx}(x/sin(x)) = \frac{sin(x)(1)-x(cos(x))}{sin^{2}(x)}$

⇒ $\frac{d}{dx}(x/sin(x)) = \frac{sin(x)-xcos(x)}{sin^{2}(x)}$

Hope this answer may help you

Answer 2

Derivatives of the following functions

1) x sinx

Here, This is product of two terms which are differentiable w.r.t x

As the product rule says,

$\sf \: (vu) ^{1} = uv ^{1} + vu ^{1}$

So,

$\frac{d}{dx} (xsinx) \\ = x \times \frac{d}{dx} (sinx) + sinx \times \frac{d}{dx} (x)$

Now sinx is a differentiable function, whose derivative is cosx.

$\frac{d}{dx} (xsinx) = x(cosx) + sinx$

2) x^4 + cosx

There are two differentiable functions as Sum.

By sum rule,

$(u + v) ^{'} = u' + v' \:$

So,

$\frac{d}{dx} ( {x}^{4} + cosx) \\ \\ = \frac{d}{dx} ( {x}^{4} ) + \frac{d}{dx} (cosx)$

Derivative of cosx is - sinx

Dedicate of x^n is nx^n-1

$\frac{d}{dx} ( {x}^{4} + cosx) \\ \\ = 4( {x}^{4 - 1} ) + ( - sinx) \\ \\ = 4 {x}^{3} - sinx$

3) x/sinx

There are two differentiable functions in division.

By Quotient rule,

$\: ( \dfrac{u}{v} )' = \dfrac{v u' - uv ' }{v ^{2} }$

$\frac{d}{dx} ( \frac{x}{sinx} ) \\ \\ = \frac{sinx \: \frac{d}{dx}(x) - x \times \frac{d}{dx}(sinx )}{ { sin}^{2}x } \\ \\ = \frac{sinx - xcosx}{sin ^{2}x }$