There is a certain function f such that the tangent line at x = 2 1 is y = 2 x + 1 .
Given that g ( x ) = f ( sin x ) , find the sum of all possible values of d x d g when evaluated at x = 6 π .
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We have that g ( x ) = f ( sin x )
Differentiate both sides with respect to x (use the chain rule on the right-hand side) to get d x d g = cos x d ( sin x ) d f ( sin x )
Now, simply evaluate the above at x = 6 π to get = 3
using the fact that the instantaneous slope of f ( x ) with respect to x at x = 2 is 2 .
slope at x = 2 1 you mean...
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g ′ ( 6 π ) = d x d g ( 6 π ) = [ d x d g ( x ) ] ( 6 π ) = [ d x d f ( s i n ( x ) ) ] ( 6 π ) = [ f ′ ( s i n ( x ) ) × d x d s i n ( x ) ] ( 6 π ) = [ f ′ ( s i n ( x ) ) × c o s ( x ) ] ( 6 π ) = f ′ ( s i n ( 6 π ) ) × c o s ( 6 π ) = f ′ ( 2 1 ) × 2 3 = 2 × 2 3 = 3
Note that f ′ ( 2 1 ) = 2 since the value of the derivative is the slope of the tangent, and the line equation given is in slope-intercept form, y ( x ) = m x + y ( 0 ) ... Alternatively, the tangent passes through the ( 0 , 1 ) and the ( 2 1 , 2 ) points and has a slope of 2 1 − 0 2 − 1 = 2 .