Applying FTC and Chain rule to calculate a minimumWhat is $g'(x)$ if $g(x) =x^2 int_x-2^sin x cos^2t dt $?Chain rule notation for function with two variablesProve that there is a number $x_0 in (0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0le x < x_0$.Chain rule for second derivativeApplying chain rule to a trace formula in matrix calculusProving the chain ruleShow that $f$ attains its minimum in $mathbb R$.Intermediate value property for derivativesFunction composition chain rule problemChain rule questions on partial derivatives
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Applying FTC and Chain rule to calculate a minimum
What is $g'(x)$ if $g(x) =x^2 int_x-2^sin x cos^2t dt $?Chain rule notation for function with two variablesProve that there is a number $x_0 in (0,1]$ such that $f(x_0)=0$ and $f(x)>0$ for $0le x < x_0$.Chain rule for second derivativeApplying chain rule to a trace formula in matrix calculusProving the chain ruleShow that $f$ attains its minimum in $mathbb R$.Intermediate value property for derivativesFunction composition chain rule problemChain rule questions on partial derivatives
$begingroup$
Find the value of $x$ where $f(x)$ attains its minimum. (Hint: you will need the Chain Rule.)
$$f(x) = int_-10^x^2+2x e^t^2,dt. $$
I'm a little confused by this. I thought this would be calculated by finding where $f'(x)=0$ by using the fundamental theorem of calculus, but the answer is $x=-1$, where $f'(-1)$ is not $0$. Any thoughts?
real-analysis
edited Mar 17 at 22:15
Robert Z
101k1070143
asked Mar 17 at 22:08
SarahSarah
141
$endgroup$
add a comment |
$begingroup$
Find the value of $x$ where $f(x)$ attains its minimum. (Hint: you will need the Chain Rule.)
$$f(x) = int_-10^x^2+2x e^t^2,dt. $$
I'm a little confused by this. I thought this would be calculated by finding where $f'(x)=0$ by using the fundamental theorem of calculus, but the answer is $x=-1$, where $f'(-1)$ is not $0$. Any thoughts?
real-analysis
edited Mar 17 at 22:15
Robert Z
101k1070143
asked Mar 17 at 22:08
SarahSarah
141
$endgroup$
add a comment |
$begingroup$
Find the value of $x$ where $f(x)$ attains its minimum. (Hint: you will need the Chain Rule.)
$$f(x) = int_-10^x^2+2x e^t^2,dt. $$
I'm a little confused by this. I thought this would be calculated by finding where $f'(x)=0$ by using the fundamental theorem of calculus, but the answer is $x=-1$, where $f'(-1)$ is not $0$. Any thoughts?
real-analysis
edited Mar 17 at 22:15
Robert Z
101k1070143
asked Mar 17 at 22:08
SarahSarah
141
$endgroup$
Find the value of $x$ where $f(x)$ attains its minimum. (Hint: you will need the Chain Rule.)
$$f(x) = int_-10^x^2+2x e^t^2,dt. $$
I'm a little confused by this. I thought this would be calculated by finding where $f'(x)=0$ by using the fundamental theorem of calculus, but the answer is $x=-1$, where $f'(-1)$ is not $0$. Any thoughts?
real-analysis
real-analysis
edited Mar 17 at 22:15
Robert Z
101k1070143
asked Mar 17 at 22:08
SarahSarah
141
edited Mar 17 at 22:15
Robert Z
101k1070143
asked Mar 17 at 22:08
SarahSarah
141
edited Mar 17 at 22:15
Robert Z
101k1070143
edited Mar 17 at 22:15
Robert Z
101k1070143
edited Mar 17 at 22:15
Robert Z
101k1070143
101k1070143
asked Mar 17 at 22:08
SarahSarah
141
asked Mar 17 at 22:08
SarahSarah
141
asked Mar 17 at 22:08
SarahSarah
141
141
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Yes, here we have to use the Fundamental Theorem of Calculus AND the Chain Rule: it follows that
$$f'(x)=e^(x^2+2x)^2cdot (x^2+2x)'=e^(x^2+2x)^2cdot 2(x+1).$$
What is $f'(-1)$? Find where $f$ is increasing/decreasing.
answered Mar 17 at 22:12
Robert ZRobert Z
101k1070143
$endgroup$
add a comment |
$begingroup$
Hint
Let $$int e^t^2dt=F(t)+C$$therefore $$f(x)=F(x^2+2x)-F(-10)$$and therefore $$f'(x)=dover dxF(x^2+2x)$$Now, applying Chain Rule leads to ...
answered Mar 17 at 22:20
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
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votes
$begingroup$
Yes, here we have to use the Fundamental Theorem of Calculus AND the Chain Rule: it follows that
$$f'(x)=e^(x^2+2x)^2cdot (x^2+2x)'=e^(x^2+2x)^2cdot 2(x+1).$$
What is $f'(-1)$? Find where $f$ is increasing/decreasing.
answered Mar 17 at 22:12
Robert ZRobert Z
101k1070143
$endgroup$
add a comment |
$begingroup$
Yes, here we have to use the Fundamental Theorem of Calculus AND the Chain Rule: it follows that
$$f'(x)=e^(x^2+2x)^2cdot (x^2+2x)'=e^(x^2+2x)^2cdot 2(x+1).$$
What is $f'(-1)$? Find where $f$ is increasing/decreasing.
answered Mar 17 at 22:12
Robert ZRobert Z
101k1070143
$endgroup$
add a comment |
$begingroup$
Yes, here we have to use the Fundamental Theorem of Calculus AND the Chain Rule: it follows that
$$f'(x)=e^(x^2+2x)^2cdot (x^2+2x)'=e^(x^2+2x)^2cdot 2(x+1).$$
What is $f'(-1)$? Find where $f$ is increasing/decreasing.
answered Mar 17 at 22:12
Robert ZRobert Z
101k1070143
$endgroup$
Yes, here we have to use the Fundamental Theorem of Calculus AND the Chain Rule: it follows that
$$f'(x)=e^(x^2+2x)^2cdot (x^2+2x)'=e^(x^2+2x)^2cdot 2(x+1).$$
What is $f'(-1)$? Find where $f$ is increasing/decreasing.
answered Mar 17 at 22:12
Robert ZRobert Z
101k1070143
edited Mar 17 at 22:17
answered Mar 17 at 22:12
Robert ZRobert Z
101k1070143
answered Mar 17 at 22:12
Robert ZRobert Z
101k1070143
answered Mar 17 at 22:12
Robert ZRobert Z
101k1070143
101k1070143
add a comment |
add a comment |
$begingroup$
Hint
Let $$int e^t^2dt=F(t)+C$$therefore $$f(x)=F(x^2+2x)-F(-10)$$and therefore $$f'(x)=dover dxF(x^2+2x)$$Now, applying Chain Rule leads to ...
answered Mar 17 at 22:20
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Let $$int e^t^2dt=F(t)+C$$therefore $$f(x)=F(x^2+2x)-F(-10)$$and therefore $$f'(x)=dover dxF(x^2+2x)$$Now, applying Chain Rule leads to ...
answered Mar 17 at 22:20
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
add a comment |
$begingroup$
Hint
Let $$int e^t^2dt=F(t)+C$$therefore $$f(x)=F(x^2+2x)-F(-10)$$and therefore $$f'(x)=dover dxF(x^2+2x)$$Now, applying Chain Rule leads to ...
answered Mar 17 at 22:20
Mostafa AyazMostafa Ayaz
18.1k31040
$endgroup$
Hint
Let $$int e^t^2dt=F(t)+C$$therefore $$f(x)=F(x^2+2x)-F(-10)$$and therefore $$f'(x)=dover dxF(x^2+2x)$$Now, applying Chain Rule leads to ...
answered Mar 17 at 22:20
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 17 at 22:20
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 17 at 22:20
Mostafa AyazMostafa Ayaz
18.1k31040
answered Mar 17 at 22:20
Mostafa AyazMostafa Ayaz
18.1k31040
18.1k31040
add a comment |
add a comment |
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