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What is the second derivative of #f(x)=cos(x^2) #?

Answer 1

Samantha Adkison

#(d^2f)/(dx^2)=-2sin(x^2)-2x^2cos(x^2)#

We use chain rule here

As #f(x)=cos(x^2)#

#(df)/(dx)=-sin(x^2)xx d/(dx)x^2=-sin(x^2)xx2x=-2xsin(x^2)#

and #(d^2f)/(dx^2)=-1*2sin(x^2)-x[cos(x^2)*2x]#

= #-2sin(x^2)-2x^2cos(x^2)#

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Answer 2

Isaiah Allen

Here,

#f(x)=cosx^2#

Diff.w.r.t., #x#, #"using "color(blue)"Chain Rule:"#

#f'(x)=-sinx^2 (d/(dx)(x^2))#

#=>f'(x)=-sinx^2(2x)#

#=>f'(x)=-2[xsinx^2]#

Again diff.w.r.t. #x#, #" using "color(blue)"product Rule :"#,

#f''(x)=-2[xcosx^2*2x+sinx^2]#

#f''(x)=-2[2x^2cosx^2+sinx^2]#

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Answer 3

Samantha Adkison

To find the second derivative of ( f(x) = \cos(x^2) ), we first find the first derivative using the chain rule, and then differentiate again using the chain rule and product rule.

The first derivative is: [ f'(x) = -2x \sin(x^2) ]

To find the second derivative, we use the product rule and chain rule: [ f''(x) = \frac{d}{dx} (-2x) \sin(x^2) + (-2x) \frac{d}{dx} \sin(x^2) ]

Simplifying and applying the chain rule for the derivative of ( \sin(x^2) ): [ f''(x) = -2 \sin(x^2) - 4x^2 \cos(x^2) ]

So, the second derivative of ( f(x) = \cos(x^2) ) is ( f''(x) = -2 \sin(x^2) - 4x^2 \cos(x^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.