Find the equation of the normal line to y = sqrt2 cosx at x= pi/4

I need the steps to do this problem along with the answer please! :)

The derivative of y (with respect to x) is

dy / dx = - sqrt(2) * sin(x).

At x = pi/4, this has the value
-sqrt(2) * sin(pi/4) = -sqrt(2) * (1/sqrt(2)) = -1.

The normal to y = sqrt2 cosx at x = pi/4 has -1 times this slope. So it has slope 1.

The normal line passes through (x,y ) = (pi/4,1) [from the equation of the curve with x = pi/4]

equation of straight line given slope, m, and a point (x0,y0) it passes through:

y-y0 = m(x-x0)

So the normal satisfies

y - 1 = 1*(x-pi/4)
=> y = x + 1 - pi/4

is the equation of the normal to y = sqrt2 cosx at x= pi/4

"I need the steps to do this problem"
I can help with that

"along with the answer please!"
But not with cheating

Remember from algebra 1, how to find a line if you know the slope and a point on the line.

Write y = mx + b with the known value of m.

Plug in the known values of x and y. Solve for b.

Well you know the slope and a point on the line.
1) Slope: Take the derivative. Substitute x = pi/4. That's the slope of the tangent. The normal line has slope -1 divided by that.

2) Point on the line. Substitute x into the expression for y.