Question

If m(x)= sin²(x), then m'(x)=? A. cos²x+sin²x. B.sinx²-cos²x C. 2cos²x-sinx D. cos²-sin²x​

Answer 1

2cos
2
x−1+2cosx=1(∵cos2x=2cos
2
x−1)
⇒2cos
2
x+2cosx−2=0
⇒cos
2
x+cosx=1.....(1)
⇒1−cos
2
x=cosx
⇒sin
2
x=cosx.....(2)

Answer 2

If `m(x) = sin²(x)`, then `m'(x)` denotes the derivative of `m(x)` with respect to `x`. We can use the chain rule and the derivative of the sine function to find `m'(x)` as follows:

```
m'(x) = d/dx(sin²(x))
= 2sin(x)*cos(x) (using the chain rule and the derivative of sin(x))
= sin(x)*2cos(x)
= 2cos(x)*sin(x)
```

Therefore, the answer is `m'(x) = 2cos(x)*sin(x)`, which is equivalent to option D, `cos²(x) - sin²(x)`.