find the derivative of cosx/sinx+cosx
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Answer:
(-1+sin³x)/sin²x
Step-by-step explanation:
d( cosx/sinx + cosx)/dx
= d(cosx/sinx)/dx + d(cosx)/dx
= {[(sinx)d(cosx)/dx - (cosx)d(sinx)/dx]/sin²x}+(-sinx)
={[(sinx)(-sinx)-(cosx)(cosx)]/sin²x} - sinx
= [-sin²x - cos²x]/sin²x} - sinx
= -1/sin²x - sinx
=-1/sin²x - sin³x/sin²x
= (-1+sin³x)/sin²x
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