Question

The derivative of the trigonometric function. Please help me.

Answer 1

`y=(sin(x^2))/(x^3)`

Apply the quotient rule:

`(\mathrm dy)/(\mathrm dx)=(x^3(\mathrm d\sin(x^2))/(\mathrm dx)-\sin(x^2)(\mathrm dx^3)/(\mathrm dx))/((x^3)^2)`

Chain and power rules:

`(\mathrm d\sin(x^2))/(\mathrm dx)=\cos(x^2)(\mathrm dx^2)/(\mathrm dx)=2x\cos(x^2)`

Power rule:

`(\mathrm dx^3)/(\mathrm dx)=3x^2`

Putting everything together, we have

`(\mathrm dy)/(\mathrm dx)=(x^3(2x\cos(x^2))-\sin(x^2)(2x\cos(x^2)))/((x^3)^2)`

`(\mathrm dy)/(\mathrm dx)=(2x^3\cos(x^2)-2x\sin(x^2)\cos(x^2))/(x^6)`

`(\mathrm dy)/(\mathrm dx)=(2x^2\cos(x^2)-\sin(2x^2))/(x^5)`

When
`x=√(\frac\pi2)`, we have

`\cos\left(\left(√(\frac\pi2)\right)^2\right)=\cos\left(\frac\pi2\right)=0`

`\sin\left(2\left(√(\frac\pi2)\right)^2\right)=\sin(\pi)=0`

so the derivative is 0.