Find the equation of the tangent line to the curve y=6tanx at the point (pi/4,6). The equation of this tangent line can be written in the form y=mx+b where m is:
and where b is:
nvm i got the answer
Good. Thanks for letting us know.
Can someone show me how this question is done?
To find the equation of the tangent line to the curve y=6tan(x) at the point (pi/4,6), we can use the property that the slope of the tangent line is equal to the derivative of the curve evaluated at that point.
Step 1: Find the derivative of the curve y=6tan(x).
The derivative of tan(x) is sec^2(x) using the chain rule. Multiply this by the constant 6 to get the derivative of y=6tan(x) as dy/dx = 6sec^2(x).
Step 2: Evaluate the derivative at the given point (pi/4,6).
Plug in pi/4 into the derivative 6sec^2(x) to get dy/dx = 6sec^2(pi/4).
Recall that the secant function of pi/4 is equal to the square root of 2. Simplify the expression to dy/dx = 6(2) = 12.
Step 3: Write the equation of the tangent line in the form y = mx + b.
We now know that the slope, m, is equal to 12. To find b, we can substitute the coordinates of the given point (pi/4,6) into the equation and solve for b.
Using the point-slope form of a line, we have:
(y - y1) = m(x - x1)
(y - 6) = 12(x - pi/4)
Expand the equation:
y - 6 = 12x - 3pi/4
Move the constant term to the other side:
y = 12x - 3pi/4 + 6
Simplify the expression to get the equation in the form y = mx + b:
y = 12x + (6 - 3pi/4)
Therefore, the equation of the tangent line is y = 12x + (6 - 3pi/4), where m = 12 and b = (6 - 3pi/4).