Question

What is the derivative of this function `sin(x^2+1)`?

Answer 1

`2xcos(x^2+1)`

Explanation:

differentiate using the `color(blue)"chain rule"`

`color(orange)"Reminder " color(red)(bar(ul(|color(white)(2/2)color(black)(d/dx[sin(f(x))]=cos(f(x)).f'(x))color(white)(22)|)))`

`rArrd/dx[sin(x^2+1)]=cos(x^2+1).d/dx(x^2+1)`

`=2xcos(x^2+1)`

Answer 2

`2xcos(x^2+1)`.

Explanation:

Let `y=sin(x^2+1)," and, let "x^2+1=t`.

`:. y=sin t, and, t=x^2+1`.

Thus, `y` is a function of `t," and, "t" of "x`.

As per the Chain Rule , then, we have,

`dy/dx=dy/(dt)(dt)/dx.............(ast)`

Now, `y=sint rArr dy/dt=cost...............(1)," and, "`

`t=x^2+1 rArr dt/dx=2x..........................(2)`.

From `(ast), (1) and (2)`, we have,

`dy/dx=2xcost=2xcos(x^2+1)`.

Enjoy Maths., and, spread the Joy!