Finding tangent line problem

Question

Find an equation of the line tangent to the curve y=6^x at point (2,36)

Answer 1

y= 6^x

y'=(ln6)6^x

y'(2)= (ln6)6^2 = 36ln6

(y-36)/(x-2)= 36ln6

y =(36ln6)x-72ln6+36

Answer 2

y' = (6^x) ln 6

Using point slope form from Algebra 1:

y - 36 = (6²) ln 6 (x - 2)

Can you take it from there and answer?

Answer 3

The derivative of 6^x is ln(6)6^x.

So the slope of the tangent line at x = 2 is y'(2) = 36ln(6).

Using point-slope:

y - 36 = 36ln(6)(x-2)

y = 36ln(6)(x - 2) +36

Check out this graph for more details:

desmos.com/calculator/vbwbo6hwno