Asked by Lee
find d/dx (sin^(21)x *cos21x)
Answers
Answered by
Steve
If that's sin(21x)*cos(21x) then
f = 1/2 sin(42x)
f' = 21cos(42x)
or, directly,
f' = 21cos(21x)cos(21x) - 21sin(21x)sin(21x)
= 21cos<sup>2</sup>21x - 21sin<sup>2</sup>21x
On the other hand, if that's
f = sin<sup>21</sup>x*cos<sup>21</sup>x, then we have
f' = 21sin<sup>20</sup>x cos<sup>22</sup>x - 21sin<sup>22</sup>x cos<sup>20</sup>x
= 21sin<sup>20</sup>x cos<sup>20</sup>x (cos<sup>2</sup>x - sin<sup>2</sup>x)
= 21sin<sup>20</sup>x cos<sup>20</sup>x cos2x
f = 1/2 sin(42x)
f' = 21cos(42x)
or, directly,
f' = 21cos(21x)cos(21x) - 21sin(21x)sin(21x)
= 21cos<sup>2</sup>21x - 21sin<sup>2</sup>21x
On the other hand, if that's
f = sin<sup>21</sup>x*cos<sup>21</sup>x, then we have
f' = 21sin<sup>20</sup>x cos<sup>22</sup>x - 21sin<sup>22</sup>x cos<sup>20</sup>x
= 21sin<sup>20</sup>x cos<sup>20</sup>x (cos<sup>2</sup>x - sin<sup>2</sup>x)
= 21sin<sup>20</sup>x cos<sup>20</sup>x cos2x
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