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Calculus Level 2

d d x [ arcsin ⁡ ( 2 x 1 + x 2 ) ] x = 1 = ? \large \dfrac{d}{dx} \Bigg[\arcsin \left( \dfrac{2x}{1+x^2} \right) \Bigg]_{x=1} = \ ?

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Surya Prakash by Surya Prakash , 22, India

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2 solutions

Kishore S. Shenoy
Oct 15, 2015

Let's take the expression right away, without making any confusion. We here use, f ( x ) = sin ⁡ − 1 ( 2 x 1 + x 2 ) = { − π − 2 tan ⁡ − 1 x x < − 1 2 tan ⁡ − 1 x x ∈ [ − 1 , 1 ] π − 2 tan ⁡ − 1 x x > 1 f(x)=\sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = \begin{cases}-\pi-2\tan^{-1}x&x<-1\\ 2\tan^{-1}x&x\in[-1,1]\\\pi-2\tan^{-1}x&x>1\end{cases}

To say that the limit exists, f ′ ( a ) = lim ⁡ x → a − f ( x ) − f ( a ) x − a = lim ⁡ x → a f ( x ) − f ( a ) x − a = lim ⁡ x → a + f ( x ) − f ( a ) x − a f'(a)=\lim_{x\to a^-}\dfrac{f(x)-f(a)}{x-a}=\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}=\lim_{x\to a^+}\dfrac{f(x)-f(a)}{x-a}

Now let's take left hand and right hand limits. lim ⁡ x → 1 + f ( x ) − f ( 1 ) x − 1 = lim ⁡ h → 0 + f ( 1 + h ) − f ( 1 ) h = lim ⁡ h → 0 + π − 2 tan ⁡ − 1 ( 1 + h ) − 2 tan ⁡ − 1 ( 1 ) h Using L’H o ˆ pital’s rule = lim ⁡ h → 0 + − 2 ( 1 + h ) 2 + 1 = − 1 lim ⁡ x → 1 − f ( x ) − f ( 1 ) x − 1 = lim ⁡ h → 0 − f ( 1 − h ) − f ( 1 ) − h = lim ⁡ h → 0 − 2 tan ⁡ − 1 ( 1 − h ) − 2 tan ⁡ − 1 ( 1 ) − h Using L’H o ˆ pital’s rule = lim ⁡ h → 0 − − 2 − 1 ⋅ [ ( 1 − h ) 2 + 1 ] = 1 ⇒ lim ⁡ x → 1 + sin ⁡ − 1 ( 2 x 1 + x 2 ) ≠ lim ⁡ x → 1 − sin ⁡ − 1 ( 2 x 1 + x 2 ) \displaystyle\begin{aligned}\lim_{x\to1^+} \dfrac{f(x)-f(1)}{x-1}&=\lim_{h\to0^+}\dfrac{f(1+h)-f(1)}h\\&=\lim_{h\to0^+}\dfrac{\pi-2\tan^{-1}(1+h)-2\tan^{-1}(1)}h\\\text{Using L'Hôpital's rule }&=\lim_{h\to0^+}\dfrac{-2}{(1+h)^2+1}\\&=\boxed{-1}\\\\\lim_{x\to1^-} \dfrac{f(x)-f(1)}{x-1}&=\lim_{h\to0^-}\dfrac{f(1-h)-f(1)}{-h}\\&=\lim_{h\to0^-}\dfrac{2\tan^{-1}(1-h)-2\tan^{-1}(1)}{-h}\\\text{Using L'Hôpital's rule }&=\lim_{h\to0^-}\dfrac{-2}{-1\cdot\left[(1-h)^2+1\right]}\\&=\boxed{1} \\\\\Rightarrow \lim_{x\to1^+}\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)&\ne \lim_{x\to1^-}\sin^{-1}\left(\dfrac{2x}{1+x^2}\right)\end{aligned}

Thus, it is not differentiable at x = 1 x = 1 , nor at x = − 1 x=-1

We can say this just by looking at the graph of sin ⁡ − 1 ( 2 x 1 + x 2 ) \sin^{-1}\left(\dfrac{2x}{1+x^2}\right)

At x = 1 x=1 , the graph has a sharp edge and thus the function is not differentiable at x = 1 x = 1

6 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Your first explanation doesn't work. You are assuming that the derivative of a function must be continuous, which is not necessarily true. There isn't a reason to evaluate lim ⁡ x → 1 f ′ ( x ) \lim_{x \rightarrow 1 } f'(x) in order to determine if f ′ ( 1 ) f'(1) exists.

Instead, what we have is that f ′ ( 1 ) = lim ⁡ x → 1 f ( x ) − f ( 1 ) x − 1 f'(1) = \lim_{x \rightarrow 1 } \frac{ f(x) - f(1) } { x - 1 } . It is this limit where the left-hand limit and the right-hand limit are different, which is why f ′ ( 1 ) f'(1) doesn't exist.

The second and third explanations work, because they are considering lim ⁡ x → 1 f ( x ) − f ( 1 ) x − 1 \lim_{x \rightarrow 1 } \frac{ f(x) - f(1) } { x - 1 }

@Calvin Lin , Yes, I thought this same thing at school today... And now you said it! Thanks Sir!

Kishore S. Shenoy - 5 years, 8 months ago

So, if we take f ′ ( 1 ) = lim ⁡ x → 1 f ( x ) − f ( 1 ) x − 1 f'(1) = \lim\limits_{x \to 1 } \frac{ f(x) - f(1) } { x - 1 } , what we get is method 3 right? So can I delete method 1 (which is wrong)?

Kishore S. Shenoy - 5 years, 8 months ago

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Actually, looking at method 3 again, it doesn't work for the same reason. What you are calculating is lim ⁡ x → 1 f ′ ( x ) \lim_{ x \rightarrow 1 } f'(x) , which may not be equal to f ′ ( 1 ) f' (1) .

And of course, method 2 via "inspection of graph" isn't a proof per se.

Calvin Lin Staff - 5 years, 8 months ago

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I can correct the third method. But correcting the first method brings me to the corrected 3rd one...

Kishore S. Shenoy - 5 years, 8 months ago

Are you able to figure out why the derivative of a differentiable function need not be continuous? If so, please add a solution

Calvin Lin Staff - 5 years, 7 months ago

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Yes! I will add a solution. Had tests these days. I'll post a solution today. Sorry for the delay...

Kishore S. Shenoy - 5 years, 7 months ago

I think that should do it @Calvin Lin sir...

Kishore S. Shenoy - 5 years, 7 months ago
Surya Prakash
Oct 8, 2015

GENERAL MISTAKE:

Take x = tan ⁡ θ x=\tan{\theta} ⟺ \iff θ = arctan ⁡ x \theta = \arctan{x} .

So, arcsin ⁡ ( 2 x 1 + x 2 ) = arcsin ⁡ ( 2 tan ⁡ θ 1 + tan ⁡ 2 θ ) = arcsin ⁡ ( sin ⁡ 2 θ ) = 2 θ = 2 arctan ⁡ x \arcsin \left(\dfrac{2x}{1+x^2} \right) = \arcsin \left( \dfrac{2 \tan{\theta}}{1+ \tan ^2 \theta} \right) = \arcsin \left( \sin 2\theta \right) = 2 \theta = 2 \arctan{x}

d d x arcsin ⁡ ( 2 x 1 + x 2 ) = d d x 2 arctan ⁡ x = 2 1 + x 2 \dfrac{d}{dx} \arcsin \left(\dfrac{2x}{1+x^2} \right) = \dfrac{d}{dx} 2 \arctan{x} = \dfrac{2}{1+x^2}

Therefore,

d d x [ arcsin ⁡ ( 2 x 1 + x 2 ) ] x = 1 = 1 \dfrac{d}{dx} \Bigg[\arcsin \left( \dfrac{2x}{1+x^2} \right) \Bigg]_{x=1} = 1

But this is actually W R O N G \color{#D61F06}{WRONG} .

CORRECT METHOD:

Let us see what happens if we bash this problem directly.

Let f ( x ) = arcsin ⁡ ( 2 x 1 + x 2 ) f(x) = \arcsin \left( \dfrac{2x}{1+x^2} \right)

f ′ ( x ) = d d x [ arcsin ⁡ ( 2 x 1 + x 2 ) ] = 1 1 − ( 2 x 1 + x 2 ) 2 × d d x ( 2 x 1 + x 2 ) = 1 + x 2 ∣ 1 − x 2 ∣ × 2 ( 1 − x 2 ) ( 1 + x 2 ) 2 \begin{aligned} f'(x) &= \dfrac{d}{dx} \Bigg[\arcsin \left( \dfrac{2x}{1+x^2} \right) \Bigg] \\ &= \dfrac{1}{\sqrt{1-\left(\dfrac{2x}{1+x^2} \right)^{2}}} \times \dfrac{d}{dx} \left(\dfrac{2x}{1+x^2} \right) \\ &= \dfrac{1+x^2}{|1-x^2|} \times \dfrac{2(1-x^2)}{(1+x^2)^2} \end{aligned}

What actually this says?? This says that the derivative of f ( x ) f(x) does not exist at x = 1 x=1 .

Be careful while calculating the derivatives using substitution method.

3 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Interesting question. Unfortunately, your analysis has several holes.

  • In General mistake , you mistakenly claim that a r c s i n ( sin ⁡ 2 θ ) = 2 θ arcsin ( \sin 2 \theta ) = 2 \theta . See Inverse Trig puzzle for why this is not true. This explains why you obtained an incorrect answer using this method.

  • Derivative using substitution method x = g ( θ ) x = g (\theta) is valid, as long as the substitution used is a differentiable function in the neighborhood, and g ′ ( θ ) ≠ 0 g'( \theta ) \neq 0 . This follows from the chain rule:

f ∘ g ( θ ) g ′ ( θ ) = f ′ ∘ g ( θ ) ) \frac{ f \circ g ( \theta ) } { g' ( \theta ) } = f' \circ g ( \theta ) )

  • The continuity of f ′ ( x ) f'(x) at x = 1 x = 1 is irrelevant to the existence of f ′ ( 1 ) f'(1) . You have not yet shown that f ′ ( 1 ) f'(1) does not exist.

  • The correct way to approach this, would be to use first principles, and show that the left-hand limit is 1, and the right-hand limit is -1, which is why the function isn't differentiable at 1 since the limits are not equal.

Interesting question, but your analysis of the problem is flawed.

Calvin Lin Staff - 5 years, 8 months ago

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is it possible to get a genralised result of differentiation of a factorial

Jaswinder Singh - 5 years, 8 months ago

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Jaswinder Singh there is general result of differentiation of factorials. Differentiation of gamma function. Γ a ( x ) = ∫ 0 ∞ ( t x − 1 ( l n t ) a e − t ) d t . { \Gamma }^{ a }\left( x \right) =\int _{ 0 }^{ \infty }{ \left( { t }^{ x-1 }{ \left( lnt \right) }^{ a }{ e }^{ -t } \right) dt. }

Aditya Kumar - 5 years, 8 months ago

I think the limit exists because at x = 1 x = 1 , arcsin ⁡ ( sin ⁡ 2 θ ) = 2 θ = 2 arctan ⁡ x \arcsin(\sin 2\theta)= 2\theta= 2\arctan x

sin ⁡ − 1 ( 2 x 1 + x 2 ) = { − π − 2 tan ⁡ − 1 x x < − 1 2 tan ⁡ − 1 x x ∈ [ − 1 , 1 ] π − 2 tan ⁡ − 1 x x > 1 \sin^{-1}\left(\dfrac{2x}{1+x^2}\right) = \begin{cases}-\pi-2\tan^{-1}x&x<-1\\ 2\tan^{-1}x&x\in[-1,1]\\\pi-2\tan^{-1}x&x>1\end{cases}

See the graph

So, the answer will be 1.

Kishore S. Shenoy - 5 years, 8 months ago

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You drew the graph of sin ⁡ − 1 2 x 1 + x 2 \sin^{-1} \frac{2x}{1+x^2}

From this, we can conclude that the limit lim ⁡ x → 1 sin ⁡ − 1 2 x 1 + x 2 \lim_{x \rightarrow 1} \sin^{-1} \frac{2x}{1+x^2} exists and is equal to π 2 \frac{\pi}{2} .

Furthermore, by looking at the slopes of the limiting tangents, the limit lim ⁡ x → 1 d d x ( sin ⁡ − 1 2 x 1 + x 2 ) \lim_{x \rightarrow 1}\frac{d}{dx} \left( \sin^{-1} \frac{2x}{1+x^2} \right) does not exist. From the LHS, the limit is 1. From the RHS, the limit is -1. Hence, the function isn't differentiable at x = 1 x = 1 .

Calvin Lin Staff - 5 years, 8 months ago

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@Calvin Lin – I get it, I get it. So, what shloud be the answer?? "Limit does not exist"??

Kishore S. Shenoy - 5 years, 8 months ago

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@Kishore S. Shenoy – I would simply go with "does not exist" or "the differential does not exist"

Can you post a solution to this problem? Thanks!

Calvin Lin Staff - 5 years, 8 months ago

sir I don't agree with you, you are suppose to simplify the expression before substitution. and I don't think that |1-x| is the same as ((1-x)^2)^1/2 like wot u assumed

Benjamin ononogbu - 5 years, 8 months ago

Little mistake. If a function is not differentiable at x = a x=a doesn't mean it is not continious at that point, for example f ( x ) = ∣ x ∣ f(x)=|x| at x = 0 x=0

Hjalmar Orellana Soto - 5 years, 8 months ago

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No it is not like that.

Aditya Kumar - 5 years, 8 months ago

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