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Find the equation of the tangent line to y=9x−3lnx at the point where x=1

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Explanation:

To find the equation of a tangent line to a function at a given point, we need to find both the slope of the tangent line (the derivative) and the y-intercept of the line.

At x = 1, the equation of the function y=9x−3lnx becomes y = 9(1) - 3ln(1) = 9.

Now let's find the derivative of the function y=9x−3lnx, which gives us the slope of the tangent line:

y' = 9 - 3(1/x)

At x = 1, the derivative y' becomes 9 - 3 = 6.

So the slope of the tangent line at x = 1 is 6.

Now we need to find the y-intercept of the line. We know that the point (1, 9) is on the line, and we know the slope is 6. Using the point-slope form of a linear equation, we can write the equation of the tangent line:

y - 9 = 6(x - 1)

Simplifying this equation gives:

y = 6x + 3

Therefore, the equation of the tangent line to y = 9x - 3ln(x) at the point where x = 1 is y = 6x + 3.

User Laurence MacNeill
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