Question

how do i do this.


find the equation for the lines that are tangent and normal to the curve y= (square root of 2)(cosx)at the point (pi/4, 1)

Answer 1

y = √2 cosx
I usually check if the given point is actually on the line
LS = 1
RS = √2 cos(pi/4) = √2(1/√2) = 1 = LS

dy/dx = -√2(sinx)

at (pi/4,1)
dy/dx = -√2(1/√2) = -1

So the slope of the tangent is -1
and the slope of the normal is +1

Take it from there,
every student taking Calculus surely has to know how to find the equation of a line given the slope and a point.

Answer 2

use product rule, then substitute value of x in dy/dx. the obtain then find equation of tangent y=mx+c.

For normal,

gradient of curve*gradient of normal=-1

find gradient of normal from equation and once again obtain normal equation in y=mx+c form