Sin 3A = 3 Sin A - 4 sin³A.
Now write the expression according to the questions requirements..
Sin3x = 3 sinx - 4 sin^3 (x)
Sin^3 x = sin3x -3 sinx ÷(- 4 )
First derivation will be like this..
The first derivatives of sin^3x = (cos3x × 3 - 3 cosx )÷ -4
Now proceed the problem for second derivative you will easily get the answer..
just use the chain rule
y = sin(3x^2)
y' = cos(3x^2)(6x)
y" = -sin(3x^2)(6x)^2 + cos(3x^2)(6)
y"' = -cos(3x^2)(6x)^3 - sin(3x^2)(2*6x)(6) - sin(3x^2)(6x)(6)
I'll let you do the algebra.
3 months ago
Now write the expression according to the questions requirements..
Sin3x = 3 sinx - 4 sin^3 (x)
Sin^3 x = sin3x -3 sinx ÷(- 4 )
First derivation will be like this..
The first derivatives of sin^3x = (cos3x × 3 - 3 cosx )÷ -4
Now proceed the problem for second derivative you will easily get the answer..