Find the derivative of xsinx by first principle. |

Hello!

Given a function `f(x)=x sin(x),` and we have to find its derivative by the definition. Consider the expression `(f(x+Delta x)-f(x))/(Delta x)` and find its limit for `Delta x->0:`

`(f(x+Delta x)-f(x))/(Delta x)=((x+Delta x)sin(x+Delta x)-x sin(x))/(Delta x)=`

`=(x(sin(x+Delta x)-sin(x))+Delta x sin(x+Delta x))/(Delta x)=`

`=x(sin(x+Delta x)-sin(x))/(Delta x)+ sin(x+Delta x).`

The second summand has the limit  `sin(x),` because `sin(x)` is continuous for any `x.`

The first summand has the limit `x(sin(x))'=xcos(x)` (by the definition of the derivative, applied...

Hello!

Given a function `f(x)=x sin(x),` and we have to find its derivative by the definition. Consider the expression `(f(x+Delta x)-f(x))/(Delta x)` and find its limit for `Delta x->0:`

`(f(x+Delta x)-f(x))/(Delta x)=((x+Delta x)sin(x+Delta x)-x sin(x))/(Delta x)=`

`=(x(sin(x+Delta x)-sin(x))+Delta x sin(x+Delta x))/(Delta x)=`

`=x(sin(x+Delta x)-sin(x))/(Delta x)+ sin(x+Delta x).`

The second summand has the limit  `sin(x),` because `sin(x)` is continuous for any `x.`

The first summand has the limit `x(sin(x))'=xcos(x)` (by the definition of the derivative, applied to `sin(x)`).

So the answer is `(x sin(x))'=x cos(x)+sin(x).`

If we "don't know" what the derivative of `sin(x)` is, transform the expression:

`(sin(x+Delta x)-sin(x))/(Delta x)=(2sin((Delta x)/2) cos(x+(Delta x)/2))/(Delta x)=`

`=sin((Delta x)/2)/((Delta x)/2) * cos(x+(Delta x)/2).`

The second factor has the limit `cos(x)` because `cos(x)` is continuous for any `x,` the first has the limit 1 (we should know this).