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  • 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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  1. anonymous
    • 4 years ago
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    Find 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: y-y1 = m(x -x1), to find the equation of your tangent

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