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anonymous
 4 years ago
Find the equation of the tangent to y = 2^(2x)  (ln 4)x + 2ln(x +1) when x=0
anonymous
 4 years ago
Find the equation of the tangent to y = 2^(2x)  (ln 4)x + 2ln(x +1) when x=0

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Find the y coordinate on the original graph when x = 0 (sub into y = 2^(2x)  (ln 4)x + 2ln(x +1)), and that point corresponds to the point on the graph (x1,y1) Now take the derivative of y = 2^(2x)  (ln 4)x + 2ln(x +1), to find y' And sub in x =0, to find the slope of the tangent (m) at the point x = 0. Now with all the information you have, you can substitute your values in the line equation: yy1 = m(x x1), to find the equation of your tangent
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