Solved

Find the Derivative of the Function g(x)=9sin(x)tanxg ( x ) = 9 \sin ( x ) \cdot \tan x

Question 62

Multiple Choice

Find the derivative of the function. g(x) =9sin(x) tanxg ( x ) = 9 \sin ( x ) \cdot \tan x


A) g(x) =9cos(x) tanx+9sin(x) csc2xg' ( x ) = 9 \cos ( x ) \cdot \tan x + 9 \sin ( x ) \cdot \csc ^ { 2 } x
B) g(x) =9cos(x) tanx+9sin(x) sec2xg '( x ) = 9 \cos ( x ) \cdot \tan x + 9 \sin ( x ) \cdot \sec ^ { 2 } x
C) g(x) =9cos(x) tanx9sin(x) cotxg '( x ) = 9 \cos ( x ) \cdot \tan x - 9 \sin ( x ) \cdot \cot x
D) g(x) =9cos(x) tanx+9cos(x) cotxg' ( x ) = 9 \cos ( x ) \cdot \tan x + 9 \cos ( x ) \cdot \cot x
E) g(x) =9cos(x) tanx9sin(x) sec2xg '( x ) = 9 \cos ( x ) \cdot \tan x - 9 \sin ( x ) \cdot \sec ^ { 2 } x

Correct Answer:

verifed

Verified

Unlock this answer now
Get Access to more Verified Answers free of charge

Related Questions