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# Find an equation of the tangent line to the curve at the given point.$y = 2^x, (0, 1)$

## $y=(\ln 2) x+1$

Derivatives

Differentiation

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Our goal here is to find the equation of the tangent line to the curve. Y equals two to the X at the points you're a one. The slope of the tangent line will be the derivative at that point, So let's find the derivative. The derivative of two to the X is two to the X Times natural log tube, so the derivative evaluated at zero will be two to the zero power types. Natural log to and to to the zero power is one. So we have one times natural log to which is natural log to So that is the slope of the tangent line. Now let's use the point slope form. Why minus y one equals m times X minus X one to find the equation of the line. So we substitute one in for why one? Since that's the Y coordinate of our point. We substitute natural log to infer the slope, and we substitute zero in for X one. Since that's the X coordinate of our point, and now we simplify. So we have y minus one equals natural log to Times X, and then we can add one to both sides. And the equation of the Tanja line is why equals the natural log of two times X plus one

Oregon State University

Derivatives

Differentiation

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