Question

find the slope-intercept form of the equation of the line tangent to the graph of y=arctan(5x) when x=√3/5

Answer 1

you need to take the derivative of atan(5x)

Answer 2

Here is a list of derivatives.
http://www.analyzemath.com/calculus/Differentiation/inverse_trigonometric.html

Answer 3

ok and then i substitute √3/5 for x of the derivative right?

Answer 4

yes, and that gives you the slope of the line

Answer 5

you then need a point on the line it will be (√3/5, y)
where y = atan(5*√3/5)

Answer 6

i got 5/4 as the final result

Answer 7

The slope of the tangent line is given by the value of the derivative at \[x=\frac{\sqrt{3}}{5}\] The derivative of \[\tan^{-1}{5x}\] is \[\frac{5}{1 + 25x^2}\] Substitue your value in for the slope and then calculate the equation of the line.

Answer 8

5/4 after substituing x in the derivative i mean

Answer 9

correct.

Answer 10

how do i get the equation? 5/4 isnt an answer...

Answer 11

you found the slope
now you need a point on the line
to find the equation of the tangent line
the point will be (√3/5, arctan(5*√3/5) )

Answer 12

then use
y - y0 = m(x -x0)

Answer 13

I think you are supposed to know arctan(sqrt(3))
but if you don't, use a calculator.... in *radian* mode.

Answer 14

5/4 is the slope of the tangent line. You also have a point on the line, namely, \[(\sqrt{3}/5, f(\sqrt{3}/5))\] where \[f(x)=\arctan{5x}\] Calculate the y value above and use the point-slope form of a line for the final answer.

Answer 15

ok thanks!