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Finding the derivative of the function $G(t)$
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Let $f$ be a differentiable and increasing , meaning that $f'(x)>0$ for all $x$ on the interval $[a,b]$, with $a>0$. Furthermore, $f$ has a differentiable inverse function $f^-1$. Let us consider the integrals
$G(t)=$$int_f(a)^f(t) pi((f^-1(y))^2-a^2) dy$ and
$H(t)=$$int_a^t 2pi x(f(t)-f(x)) dx$.
Show that $G(t)$ and $H(t)$ have identical derivatives on $[a,b]$. I think we must use Leibniz integral rule here, but how can I apply it to $G(t)$? In case of $H(t)$ it is quite easier than that to apply on $G(t)$. PLease help me to find the derivative of $G(t)$.
derivatives
asked Jul 17 at 6:12
abcdmath
29310
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up vote
-1
down vote
favorite
Let $f$ be a differentiable and increasing , meaning that $f'(x)>0$ for all $x$ on the interval $[a,b]$, with $a>0$. Furthermore, $f$ has a differentiable inverse function $f^-1$. Let us consider the integrals
$G(t)=$$int_f(a)^f(t) pi((f^-1(y))^2-a^2) dy$ and
$H(t)=$$int_a^t 2pi x(f(t)-f(x)) dx$.
Show that $G(t)$ and $H(t)$ have identical derivatives on $[a,b]$. I think we must use Leibniz integral rule here, but how can I apply it to $G(t)$? In case of $H(t)$ it is quite easier than that to apply on $G(t)$. PLease help me to find the derivative of $G(t)$.
derivatives
asked Jul 17 at 6:12
abcdmath
29310
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Let $f$ be a differentiable and increasing , meaning that $f'(x)>0$ for all $x$ on the interval $[a,b]$, with $a>0$. Furthermore, $f$ has a differentiable inverse function $f^-1$. Let us consider the integrals
$G(t)=$$int_f(a)^f(t) pi((f^-1(y))^2-a^2) dy$ and
$H(t)=$$int_a^t 2pi x(f(t)-f(x)) dx$.
Show that $G(t)$ and $H(t)$ have identical derivatives on $[a,b]$. I think we must use Leibniz integral rule here, but how can I apply it to $G(t)$? In case of $H(t)$ it is quite easier than that to apply on $G(t)$. PLease help me to find the derivative of $G(t)$.
derivatives
asked Jul 17 at 6:12
abcdmath
29310
Let $f$ be a differentiable and increasing , meaning that $f'(x)>0$ for all $x$ on the interval $[a,b]$, with $a>0$. Furthermore, $f$ has a differentiable inverse function $f^-1$. Let us consider the integrals
$G(t)=$$int_f(a)^f(t) pi((f^-1(y))^2-a^2) dy$ and
$H(t)=$$int_a^t 2pi x(f(t)-f(x)) dx$.
Show that $G(t)$ and $H(t)$ have identical derivatives on $[a,b]$. I think we must use Leibniz integral rule here, but how can I apply it to $G(t)$? In case of $H(t)$ it is quite easier than that to apply on $G(t)$. PLease help me to find the derivative of $G(t)$.
derivatives
asked Jul 17 at 6:12
abcdmath
29310
asked Jul 17 at 6:12
abcdmath
29310
asked Jul 17 at 6:12
abcdmath
29310
asked Jul 17 at 6:12
abcdmath
29310
29310
add a comment |Â
add a comment |Â
1 Answer
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$G'(t)=pi [(f^-1f(t))^2-a^2]f'(t)=pi [t^2-a^2]f'(t)$ by chain rule. $H(t)=f(t)int_a^t 2pi x , dx-int_a^t2pi xf(x), dx$ so $H'(t)=2pi tf(t)+pi (t^2-a^2) f'(t)-2pi f(t)=pi (t^2-a^2) f'(t)$. Hence $H'(t)=G'(t)$.
answered Jul 17 at 6:37
Kavi Rama Murthy
21.1k2829
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$G'(t)=pi [(f^-1f(t))^2-a^2]f'(t)=pi [t^2-a^2]f'(t)$ by chain rule. $H(t)=f(t)int_a^t 2pi x , dx-int_a^t2pi xf(x), dx$ so $H'(t)=2pi tf(t)+pi (t^2-a^2) f'(t)-2pi f(t)=pi (t^2-a^2) f'(t)$. Hence $H'(t)=G'(t)$.
answered Jul 17 at 6:37
Kavi Rama Murthy
21.1k2829
add a comment |Â
up vote
1
down vote
$G'(t)=pi [(f^-1f(t))^2-a^2]f'(t)=pi [t^2-a^2]f'(t)$ by chain rule. $H(t)=f(t)int_a^t 2pi x , dx-int_a^t2pi xf(x), dx$ so $H'(t)=2pi tf(t)+pi (t^2-a^2) f'(t)-2pi f(t)=pi (t^2-a^2) f'(t)$. Hence $H'(t)=G'(t)$.
answered Jul 17 at 6:37
Kavi Rama Murthy
21.1k2829
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$G'(t)=pi [(f^-1f(t))^2-a^2]f'(t)=pi [t^2-a^2]f'(t)$ by chain rule. $H(t)=f(t)int_a^t 2pi x , dx-int_a^t2pi xf(x), dx$ so $H'(t)=2pi tf(t)+pi (t^2-a^2) f'(t)-2pi f(t)=pi (t^2-a^2) f'(t)$. Hence $H'(t)=G'(t)$.
answered Jul 17 at 6:37
Kavi Rama Murthy
21.1k2829
$G'(t)=pi [(f^-1f(t))^2-a^2]f'(t)=pi [t^2-a^2]f'(t)$ by chain rule. $H(t)=f(t)int_a^t 2pi x , dx-int_a^t2pi xf(x), dx$ so $H'(t)=2pi tf(t)+pi (t^2-a^2) f'(t)-2pi f(t)=pi (t^2-a^2) f'(t)$. Hence $H'(t)=G'(t)$.
answered Jul 17 at 6:37
Kavi Rama Murthy
21.1k2829
answered Jul 17 at 6:37
Kavi Rama Murthy
21.1k2829
answered Jul 17 at 6:37
Kavi Rama Murthy
21.1k2829
answered Jul 17 at 6:37
Kavi Rama Murthy
21.1k2829
21.1k2829
add a comment |Â
add a comment |Â
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