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Question
If y = tanx + secx, prove that `("d"^2y)/("d"x^2) = cosx/(1 - sinx)^2`
Solution
We have y = tanx + secx.
Differentiating w.r.t. x, we get
`("d"y)/("d"x)` = sec2x + secx tanx
= `1/(cos2x) + sinx/(cos^2x)`
= (1 + sinx)/(cos^2x)`
= `(1 + sinx)/((1 + sinx)(1 - sinx))`
Thus `("d"y)/("d"x) = 1/(1 - sin )`.
Now, differentiating again w.r.t. x, we get
`("d"^2y)/("d"x^2) = (-(cos x))/(1 - sin x)^2`
= `cosx/(1 - sin x)^2`
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