Question

Let \(f(x) = \int_{2}^{x} \frac{dt}{\sqrt{1+t^4}} \) and \(g\) be the the inverse of \(f\). Then, the value of \(g'(0)\) is ... ?

Answer 1

Let \(f(x) = \int_{2}^{x} \frac{dt}{\sqrt{1+t^4}} \) and \(g\) be the the inverse of \(f\). Then, the value of \(g'(0)\) is ... ?

Answer 2

Hint : (2, 0) is a point on f(x)

Answer 3

I wished to follow the conventional way. That is to first find \(f\) and then find it's inverse and carry on this way.

Here is what I did till now:

I did a substitution of \(t^2 = \tan \theta\) which gives : \(2t dt = \sec ^2 \theta \implies dt = \cfrac{\sec^2 \theta}{\sqrt{\tan \theta}} \)

The integral becomes:

\[\int_{2}^{x} \cfrac{\sec \theta}{2\sqrt{\tan \theta}} d \theta \]

Should I follow the product rule or another substitution?

Answer 4

clearly thats very painful, we don't need to find g(x),
the question is simply about finding g'(0)

Answer 5

Can you just show me how does that hint help here? And how did you arrive to that?

Answer 6

My book found \(f'(x)\) first and then found \(g'(x)\).

I'm not sure how did it do that though.

Answer 7

In light of the earlier hint, start by finding \(f'(x)\) :
\[f(x) = \int_{2}^{x} \frac{dt}{\sqrt{1+t^4}} \implies f'(x) = \frac{1}{\sqrt{1+x^4}}\]

Answer 8

Yeah, that's what my book shows! But, am very unsure how did you arrive to \(f'(x)\) .

Why didn't we consider the lower limit 2 ? Why only x?

Answer 9

lets work that in steps

Answer 10

\[f(x) = \int_{2}^{x} \frac{dt}{\sqrt{1+t^4}} \]
let \(F(t) = \int \frac{dt}{\sqrt{1+t^4}}\implies F'(t) = \frac{1}{\sqrt{1+t^4}} \)

yes ?

Answer 11

Yeah!

Answer 12

Next apply fundamental theorem of calculus :
\[f(x) = \int_{2}^{x} \frac{dt}{\sqrt{1+t^4}} = F(x)-F(2) \]

still yes ?

Answer 13

Ahh.. I see, and derivative of \(F(2)\) will equal zero.

Answer 14

Exactly!

Answer 15

That's awesome. So, next we will find \(g'(x)\) . Just a second, I'll give it a try.

Answer 16

k

Answer 17

What does this mean?