Find the gradient of curve y=xtanx at x=π/4

I thought it was π/2, but my book says it is π/2 +1

Using the product rule we get:

dy/dx = (1)tanx + x(sec²x)

so, when x = π/4 we get:

dy/dx = tan(π/4) + (π/4)(sec²(π/4))

Now, sec²(π/4) = 1/cos²(π/4) => 1/(1/√2)² = 2

=> dy/dx = 1 + 2(π/4)...i.e. 1 + π/2

:)>

y=xtan(x)
dy/dx = tan(x) d/dx 1 + x d/dx sec^2(x)
= tan(x) + xsec^2(x)

when x = π/4

= tan(π/4) + π/4sec^2(π/4)
= 1 + π/4√(2)^2
= 1 + π/4 (2)
= 1 + π/2 answer//

dy/dx = tanx + xsec²x . . . using the product rule
Substitute x = ¼π
dy/dx = tan¼π + ¼πsec² ¼π
dy/dx = 1 + ¼π / (½√2)²
dy/dx = 1 + ¼π/½
dy/dx = 1 + ½π