Expressing defined integral as composition of differentiable functions

[lep11]

Homework Statement

Let $f(t)=\int_{t}^{t^2} \frac{1}{s+\sin{s}}ds,t>1.$Express $f$ as a composition of two differentiable functions $g:ℝ→ℝ^2$ and $h:ℝ^2→ℝ$. In addition, find the derivative of $f$ (using the composition).

Homework Equations

The Attempt at a Solution

Honestly, I have no proper idea how to approach this problem. I know what is being asked, but how to find such functions $g$ and $f$? Let g(t)=(t,sint) and h(x,y)=1/(x+y)? A nudge in the right direction will be appreciated.

 

[PeroK]

Homework Statement

Let $f(t)=\int_{t}^{t^2} \frac{1}{s+\sin{s}}ds,t>1.$Express $f$ as a composition of two differentiable functions $g:ℝ→ℝ^2$ and $h:ℝ^2→ℝ$. In addition, find the derivative of $f$ (using the composition).

Homework Equations

The Attempt at a Solution

Honestly, I have no proper idea how to approach this problem. I know what is being asked, but how to find such functions $g$ and $f$? Let g(t)=(t,sint) and h(x,y)=1/(x+y)? A nudge in the right direction will be appreciated.

What about $g(t) = (t, t^2)$?

PS I think I can see what is intended, but if the aim is to find $f'$ I don't see why you need to consider functions of more than one variable.

 

[lep11]

What about $g(t) = (t, t^2)$?

PS I think I can see what is intended, but if the aim is to find $f'$ I don't see why you need to consider functions of more than one variable.

The problem actually consists of two parts (a and b) and the part b is to find the derivative of $f$ using the composition. $g(t)=(t,t^2)$, well, okay, but how to find such $h$? $h(x,y)=\int_{x}^{y} \frac{1}{s+\sin{s}}ds$? And in part b in thinking maybe it's the reversal chain rule? It's a bit confusing.

 

[PeroK]

The problem actually consists of two parts (a and b) and the part b is to find the derivative of $f$ using the composition. $g(t)=(t,t^2)$, well, okay, but how to find such $h$? $h(x,y)=\int_{x}^{y} \frac{1}{s+\sin{s}}ds$?

Yes, $f(t) = h(g(t))$ with those definitions.

For your information, I would have split $f$ as follows:

$f(t) = f_1(t^2) - f_1(t)$

Where $f_1(t) = \int_0^{t} k(s)ds$

Those are all differentiable functions of a single variable.

 

[lep11]

Yes, $f(t) = h(g(t))$ with those definitions.

Are there other possibilities?

 

[PeroK]

Are there other possibilities?

There may be, but I think that is the one that the question setter intended.

 

[lep11]

By the chain rule $\frac{\partial{f}}{\partial{t}}(t)=\frac{\partial{h(g(t))}}{\partial{t}}=\frac{\partial{h}}{\partial{g_1}}\frac{\partial{g_1}}{\partial{t}}+\frac{\partial{h}}{\partial{g_2}}\frac{\partial{g_2}}{\partial{t}}$

How to apply it in this case?

 

[PeroK]

By the chain rule $\frac{\partial{f}}{\partial{t}}(t)=\frac{\partial{h(g(t))}}{\partial{t}}=\frac{\partial{h}}{\partial{g_1}}\frac{\partial{g_1}}{\partial{t}}+\frac{\partial{h}}{\partial{g_2}}\frac{\partial{g_2}}{\partial{t}}$

How to apply it in this case?

First, $f$ is a function of $t$, so you can't have a partial derivative.

You need to write:

$f(t) = g(x(t), y(t)) = g(t, t^2)$

And, apply the chain rule to that.