Question

If y =sin x sec (90-x) , what is value of dy/dx of the previous ?

Answer 1

We can use the chain rule to find the derivative of y with respect to x.

The chain rule states that if y = f(u) and u = g(x), then dy/dx = dy/du * du/dx

y =sin x sec (90-x)

dy/dx = cos x sec (90-x) * d/dx(sec (90-x)) - sin x tan (90-x) sec (90-x) * d/dx(90-x)

let us calculate the second derivative

du/dx = -cos(90-x)

d²y/dx² = -sin x sec (90-x) * cos (90-x) - cos x tan (90-x) sec (90-x) * cos (90-x) - cos x sec (90-x) * (-sin (90-x))

= -sin x sec (90-x) * cos (90-x) - cos x tan (90-x) sec (90-x) * cos (90-x) + cos x sec (90-x) * sin (90-x)

= -sin x sec (90-x) * cos (90-x) + cos x tan (90-x) sec (90-x) * sin (90-x)

= -(sin^2 x + cos^2 x) sec (90-x) tan (90-x)

= -sec^2 (90-x) tan (90-x)

So, the value of the second derivative of y with respect to x is -sec^2 (90-x) tan (90-x)

Explanation: