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What is the second derivative of f(x)=secx?

Answer
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Hint: Here, the given question has a trigonometric function. We have to find the derivative or differentiated term of the function. We know the standard derivative of secant function and which is the first derivative. To find the second derivative we use product rule that is dydx=u×dvdx+v×dudx.

Complete step by step solution:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Given f(x)=secx.
f(x)=1cosx
Or
f(x)=(cosx)1
Now differentiate with respect to ‘x’, we have,
f(x)=ddx(cosx)1
f(x)=(cosx)11ddx(cosx)
f(x)=(cosx)2(sinx)
f(x)=sinx(cosx)2
f(x)=sinxcosx1cosx
f(x)=tanx.secx. This is the first derivative.
To find the second derivative we again differentiate the first derivative.
f(x)=ddx(tanx.secx)
Using the product rule of differentiation, we have,
f(x)=tanxddx(secx)+secxddx(tanx).
We know ddx(secx)=tanx.secx and ddx(tanx)=sec2x,
f(x)=tanx(secx.tanx)+secx(sec2x)
f(x)=secx.tan2x+sec3x. This is the required result.
Thus the second derivative of f(x)=secx is secx.tan2x+sec3x.

Additional information:
Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as h(x)=af(x)+bg(x)
Quotient rule: The derivative of one function divided by other is found by quotient rule such as[f(x)g(x)]=g(x)f(x)f(x)g(x)[g(x)]2.
Product rule: When a derivative of a product of two function is to be found, then we use product rule that is dydx=u×dvdx+v×dudx.
Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is fog(x0)=[(fog)(x0)]g(x0).

Note:
We know the differentiation of xn is d(xn)dx=n.xn1. The first obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.