Question

Differentiate sin(ax+b)*cos(CX+d)?

Answer 1

`acos(ax+b)cos(cx+d)−csin(ax+b)sin(cx+d)`

Explanation:

`d/dx[sin(ax+b)cos(cx+d)]d/dx[sin(ax+b)cos(cx+d)] =d/dx[sin(ax+b)]⋅cos(cx+d)+sin(ax+b)⋅d/dx[cos(cx+d)] =d/dx[sin(ax+b)]⋅cos(cx+d)+sin(ax+b)⋅d/dx[cos(cx+d)] =cos(ax+b)⋅d/dx[ax+b]⋅cos(cx+d)+(−sin(cx+d))⋅d/dx[cx+d]⋅sin(ax+b) =cos⁡(ax+b)⋅d/dx[ax+b]⋅cos(cx+d)+(−sin(cx+d))⋅d/dx[cx+d]⋅sin(ax+b) =(a⋅d/dx[x]+d/dx[b])cos(ax+b)cos(cx+d)−(c⋅d/dx[x]+d/dx[d])sin(ax+b)sin(cx+d) =(a⋅d/dx[x]+d/dx[b])cos(ax+b)cos(cx+d)−(c⋅d/dx[x]+d/dx[d])sin(ax+b)sin(cx+d) =(1a+0)cos(ax+b)cos(cx+d)−(1c+0)sin(ax+b)sin(cx+d) =(1a+0)cos(ax+b)cos(cx+d)−(1c+0)sin(ax+b)sin(cx+d) =acos(ax+b)cos(cx+d)−csin(ax+b)sin(cx+d)`

Answer 2

Differential of `sin(ax+b)cos(cx+d)` is `acos(cx+d)cos(ax+b)-csin(ax+b)sin(cx+d))`

Explanation:

We use product formula here, which states that if `y=g(x)h(x)`, then `(dy)/(dx)=g(x)(dh(x))/(dx)+h(x)(dg(x))/(dx)`

Hence for differential of `y=sin(ax+b)cos(cx+d)`, observe that

we have `g(x)=sin(ax+b)` and `h(x)=cos(cx+d)`

therefore `(dg(x))/(dx)=acos(ax+b)` and `(dh(x))/(dx)=-csin(cx+d)`

Hence differential of `sin(ax+b)cos(cx+d)` is

`sin(ax+b)xx(-csin(cx+d))+cos(cx+d)xxacos(ax+b)`

= `acos(cx+d)cos(ax+b)-csin(ax+b)sin(cx+d))`

Answer 3

`1/2[(a+c)cos{(a+c)x+(b+d)}`

`+(a-c)cos{(a-c)x+(b-d)}]`.

Explanation:

Recall that, `2sinucosv=sin(u+v)+sin(u-v)`.

Let, `y=sin(ax+b)cos(cx+d)`.

`:. 2y=2sin(ax+b)cos(cx+d)`,

`=sin(ax+b+cx+d)+sin(ax+b-cx-d`,

`=sin{(a+c)x+(b+d)}+sin{(a-c)x+(b-d)}`.

`:. d/dx(2y)=d/dx[sin{(a+c)x+(b+d)}`

`+sin{(a-c)x+(b-d)}]`.

`:. 2dy/dx=cos{(a+c)x+(b+d)}*d/dx{(a+c)x+(b+d)}`

`+cos{(a-c)x+(b-d)}*d/dx{(a-c)x+(b-d)}`,

`=(a+c)cos{(a+c)x+(b+d)}`

`+(a-c)cos{(a-c)x+(b-d)}`.

`rArr dy/dx=1/2[(a+c)cos{(a+c)x+(b+d)}`

`+(a-c)cos{(a-c)x+(b-d)}]`.