Question

find the equation of the tangent line at the point specified..
y = xtanx ; x = pi/4

Answer 1

x*(secx)^2 + tan(x), plug in x = pi/4

Answer 2

take the derivative, using the product rule.
replace x by
\[\frac{\pi}{4}\] then use point slope formula

Answer 3

yes but do i have to apply the power rule twice, once with (tanx) with x(tanx)?

Answer 4

no pwer rules, just product rule

Answer 5

sorry thats what i meant. do i have to apply it twice?

Answer 6

no, just once. This is in form of u*v, where u = x and v = tanx
(u*v)' = u*v' + v*u' = x*(Secx)^2 + (tanx)*1

Answer 7

ok thanks

Answer 8

y = xtanx ; x = pi/4
f'(x) = tanx + x ( sec^2x)
= tanx + x( 1 + tan^2x)
-> f'( pi/4) = tan(pi/4) + pi/4 ( 1 + tan^2pi/4)
= 1 + pi/4 + pi/4 = 1 + pi/2 = 2.57
Tangent line:
y - pi/4 = 2.57( x - pi/4)