Question

Differentiate `xsinx` using first principles?

Answer 1

`f'(x)=xcosx+sinx`

Explanation:

Derivation from first principles tells us that for a function `f(x)`, `f'(x)=lim_(h->0)(f(x+h)-f(x))/h`

In this case, `f(x)=xsinx`, so we have:
`f'(x)=lim_(h->0)((x+h)sin(x+h)-xsinx)/h`

We can use the identity `sin(A+B)=sinAcosB+sinBcosA`

`color(white)(f'(x))=lim_(h->0)((x+h)(sin(x)cos(h)+cos(x)sin(h))-xsinx)/h`

`color(white)(f'(x))=lim_(h->0)(xsin(x)cos(h)+xcos(x)sin(h)+hsin(x)cos(h)+hcos(x)sin(h)-xsinx)/h`

`color(white)(f'(x))=lim_(h->0)((xsin(x)cos(h))/h+(xcos(x)sin(h))/h+(hsin(x)cos(h))/h+(hcos(x)sin(h))/h-(xsinx)/h)`

`color(white)(f'(x))=lim_(h->0)((xsin(x)cos(h))/h+(xcos(x)sin(h))/h+sin(x)cos(h)+cos(x)sin(h)-(xsinx)/h)`

It is known that:
`lim_(h->0)sinh/h=1`

`color(white)(f'(x))=lim_(h->0)((xsin(x)cos(h))/h+xcos(x)+sin(x)cos(h)+cos(x)sin(h)-(xsinx)/h)`

`color(white)(f'(x))=xcos(x)+sin(x)cos(0)+cos(x)sin(0)+lim_(h->0)((xsin(x)cos(h))/h-(xsinx)/h)`

`color(white)(f'(x))=xcos(x)+sin(x)+lim_(h->0)((xsin(x)cos(h))/h-(xsinx)/h)`

`color(white)(f'(x))=xcos(x)+sin(x)+lim_(h->0)((xsin(x)cos(h)-xsinx)/h)`

`color(white)(f'(x))=xcos(x)+sin(x)+lim_(h->0)((xsin(x)(cos(h)-1))/h)`

`color(white)(f'(x))=xcos(x)+sin(x)+xsin(x)lim_(h->0)(cos(h)-1)/h`

It is also known that:
`lim_(h->0)(cos(h)-1)/h=0`

`color(white)(f'(x))=xcos(x)+sin(x)+xsin(x)0`

`color(white)(f'(x))=xcos(x)+sin(x)`

Proof:
`f(x)=xsinx`

`f'(x)=d/dx[x]sinx+xd/dx[sinx]`
`color(white)(f'(x))=1sinx+xcosx`
`color(white)(f'(x))=xcosx+sinx`

Answer 2

`d/dx xsinx = x cosx +sinx`

Explanation:

Let us define:

`f(x) = xsinx`

Using the limit definition of the derivative, we compute the derivative using:

`f'(x) = lim_(h rarr 0) (f(x+h)-f(x))/h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) ((x+h)sin(x+h) - xsinx)/h`

`\ \ \ \ \ \ \ \ \ = lim_(h rarr 0) {x((sin(x+h) - sinx))/h +sin(x+h)}`

`\ \ \ \ \ \ \ \ \ = x lim_(h rarr 0) ((sin(x+h) - sinx))/h +lim_(h rarr 0)sin(x+h)`

`\ \ \ \ \ \ \ \ \ = x {lim_(h rarr 0) (sin(x+h) - sinx)/h} +sinx`

So we only need to calculate the limit:

`L = lim_(h rarr 0) (sin(x+h) - sinx)/h`

Which, coincidentally, is the limit arising from the derivative of `sinx` (by definition). Now using the trigonometric multiply angle identity:

`sin(A+B) -= sinA cosB+cosA sinB`

We get:

`L = lim_(h rarr 0) (sin x cos h + cos x sin h - sin x)/h`

`\ \ \ = lim_(h rarr 0) (sin x (cos h -1))/h+ (cos x sin h)/h`

`\ \ \ = sinx \ {lim_(h rarr 0) (cos h -1)/h } + cos x \ {lim_(h rarr 0) ( sin h)/h}`

Both of these limits are standard calculus limits, and providing `x` is in radians, we have:

`lim_(h rarr 0) (cos h -1)/h =0 \ \` and `\ \ lim_(h rarr 0) ( sin h)/h = 1`

With these results, we get the result:

`L = cos x => f'(x) = x cosx +sinx`