Home > Calculus > Relationship between First and Second Derivatives of a Function

If #f(x)=sec(x)#, how do I find #f''(π/4)#?

Answer 1

Cameron Alexander

First find the expression for #f''(x)#, then evaluate it at #pi/4#.

#f(x)=secx#.

#f'(x)=secxtanx#.

Use the product rule to find #f''(x)#.

#f''(x)=(secxtanx)tanx+secx(sec^2x)#.

So, #f''(x)=secxtan^2x+sec^3x#.

Evaluate:

#f''(pi/4)=sec(pi/4)tan^2(pi/4)+sec^3(pi/4)#

#=(sqrt2)(1)^2+(sqrt2)^3=sqrt2+2sqrt2=3sqrt2#.

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Answer 2

Axel Alvarez

To find ( f''(\frac{\pi}{4}) ) for ( f(x) = \sec(x) ), you first need to find the first and second derivatives of ( \sec(x) ) with respect to ( x ).

The first derivative of ( \sec(x) ) is ( \sec(x) \tan(x) ).

The second derivative of ( \sec(x) ) is ( \sec(x) \tan(x) \tan(x) + \sec(x) \sec^2(x) = \sec(x)(\tan^2(x) + \sec^2(x)) ).

Now, evaluate ( f''(\frac{\pi}{4}) ) by substituting ( \frac{\pi}{4} ) into the second derivative expression.

( \sec(\frac{\pi}{4}) = \sqrt{2} ) and ( \tan(\frac{\pi}{4}) = 1 ).

So, ( f''(\frac{\pi}{4}) = \sqrt{2}(\tan^2(\frac{\pi}{4}) + \sec^2(\frac{\pi}{4})) = \sqrt{2}(1 + 2) = \sqrt{2}(3) = 3\sqrt{2} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.