Find an equation of the tangent line of the exponential function at the point (0,1)
$begingroup$
So I differentiated the expression
$$ y=e^{2x}cos(pi x) $$
And I got
$$ y'=2e^{2x}cos(pi x)-pi e^{2x}sin(pi x)$$
But when from the given point $(0,1)$ when I plug in zero I get 2
I looked the answer up in the back of my textbook (which I know I shouldn't do) and the answer is $y=2x+1$
calculus exponential-function
edited Dec 11 '18 at 2:33
Dando18
4,68241235
asked Dec 11 '18 at 2:26
Eric BrownEric Brown
757
$endgroup$
add a comment |
$begingroup$
So I differentiated the expression
$$ y=e^{2x}cos(pi x) $$
And I got
$$ y'=2e^{2x}cos(pi x)-pi e^{2x}sin(pi x)$$
But when from the given point $(0,1)$ when I plug in zero I get 2
I looked the answer up in the back of my textbook (which I know I shouldn't do) and the answer is $y=2x+1$
calculus exponential-function
edited Dec 11 '18 at 2:33
Dando18
4,68241235
asked Dec 11 '18 at 2:26
Eric BrownEric Brown
757
$endgroup$
$begingroup$
$y'(0)=2$ gives you the slope of the line at $(0,1)$. You still need to find the equation of the line (given it's slope and a point lying on it).
$endgroup$
– Anurag A
Dec 11 '18 at 2:28
$begingroup$
I think this might be a case where you're not sure what you're actually looking for. You say you get $2$, which is indeed $y'(0)$. But you're not looking for a number, you're looking for a line.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:30
add a comment |
$begingroup$
So I differentiated the expression
$$ y=e^{2x}cos(pi x) $$
And I got
$$ y'=2e^{2x}cos(pi x)-pi e^{2x}sin(pi x)$$
But when from the given point $(0,1)$ when I plug in zero I get 2
I looked the answer up in the back of my textbook (which I know I shouldn't do) and the answer is $y=2x+1$
calculus exponential-function
edited Dec 11 '18 at 2:33
Dando18
4,68241235
asked Dec 11 '18 at 2:26
Eric BrownEric Brown
757
$endgroup$
So I differentiated the expression
$$ y=e^{2x}cos(pi x) $$
And I got
$$ y'=2e^{2x}cos(pi x)-pi e^{2x}sin(pi x)$$
But when from the given point $(0,1)$ when I plug in zero I get 2
I looked the answer up in the back of my textbook (which I know I shouldn't do) and the answer is $y=2x+1$
calculus exponential-function
calculus exponential-function
edited Dec 11 '18 at 2:33
Dando18
4,68241235
asked Dec 11 '18 at 2:26
Eric BrownEric Brown
757
edited Dec 11 '18 at 2:33
Dando18
4,68241235
asked Dec 11 '18 at 2:26
Eric BrownEric Brown
757
edited Dec 11 '18 at 2:33
Dando18
4,68241235
edited Dec 11 '18 at 2:33
Dando18
4,68241235
edited Dec 11 '18 at 2:33
Dando18
4,68241235
4,68241235
asked Dec 11 '18 at 2:26
Eric BrownEric Brown
757
asked Dec 11 '18 at 2:26
Eric BrownEric Brown
757
asked Dec 11 '18 at 2:26
Eric BrownEric Brown
757
757
$begingroup$
$y'(0)=2$ gives you the slope of the line at $(0,1)$. You still need to find the equation of the line (given it's slope and a point lying on it).
$endgroup$
– Anurag A
Dec 11 '18 at 2:28
$begingroup$
I think this might be a case where you're not sure what you're actually looking for. You say you get $2$, which is indeed $y'(0)$. But you're not looking for a number, you're looking for a line.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:30
add a comment |
$begingroup$
$y'(0)=2$ gives you the slope of the line at $(0,1)$. You still need to find the equation of the line (given it's slope and a point lying on it).
$endgroup$
– Anurag A
Dec 11 '18 at 2:28
$begingroup$
I think this might be a case where you're not sure what you're actually looking for. You say you get $2$, which is indeed $y'(0)$. But you're not looking for a number, you're looking for a line.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:30
$begingroup$
$y'(0)=2$ gives you the slope of the line at $(0,1)$. You still need to find the equation of the line (given it's slope and a point lying on it).
$endgroup$
– Anurag A
Dec 11 '18 at 2:28
$begingroup$
$y'(0)=2$ gives you the slope of the line at $(0,1)$. You still need to find the equation of the line (given it's slope and a point lying on it).
$endgroup$
– Anurag A
Dec 11 '18 at 2:28
$begingroup$
I think this might be a case where you're not sure what you're actually looking for. You say you get $2$, which is indeed $y'(0)$. But you're not looking for a number, you're looking for a line.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:30
$begingroup$
I think this might be a case where you're not sure what you're actually looking for. You say you get $2$, which is indeed $y'(0)$. But you're not looking for a number, you're looking for a line.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:30
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are looking for a line $(y-y_0)=m(x-x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope of the line.
You are correct in finding the derivative, since $m=y'(0)$. That is, the derivative at $x_0$ is the slope of the tangent line at $x_0$. So now you just have to use the point given to find the equation of the line,
$$ (y-1) = 2(x-0) implies y=2x+1 $$
answered Dec 11 '18 at 2:37
Dando18Dando18
4,68241235
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
You are looking for a line $(y-y_0)=m(x-x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope of the line.
You are correct in finding the derivative, since $m=y'(0)$. That is, the derivative at $x_0$ is the slope of the tangent line at $x_0$. So now you just have to use the point given to find the equation of the line,
$$ (y-1) = 2(x-0) implies y=2x+1 $$
answered Dec 11 '18 at 2:37
Dando18Dando18
4,68241235
$endgroup$
add a comment |
$begingroup$
You are looking for a line $(y-y_0)=m(x-x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope of the line.
You are correct in finding the derivative, since $m=y'(0)$. That is, the derivative at $x_0$ is the slope of the tangent line at $x_0$. So now you just have to use the point given to find the equation of the line,
$$ (y-1) = 2(x-0) implies y=2x+1 $$
answered Dec 11 '18 at 2:37
Dando18Dando18
4,68241235
$endgroup$
add a comment |
$begingroup$
You are looking for a line $(y-y_0)=m(x-x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope of the line.
You are correct in finding the derivative, since $m=y'(0)$. That is, the derivative at $x_0$ is the slope of the tangent line at $x_0$. So now you just have to use the point given to find the equation of the line,
$$ (y-1) = 2(x-0) implies y=2x+1 $$
answered Dec 11 '18 at 2:37
Dando18Dando18
4,68241235
$endgroup$
You are looking for a line $(y-y_0)=m(x-x_0)$, where $(x_0,y_0)$ is a point on the line and $m$ is the slope of the line.
You are correct in finding the derivative, since $m=y'(0)$. That is, the derivative at $x_0$ is the slope of the tangent line at $x_0$. So now you just have to use the point given to find the equation of the line,
$$ (y-1) = 2(x-0) implies y=2x+1 $$
answered Dec 11 '18 at 2:37
Dando18Dando18
4,68241235
answered Dec 11 '18 at 2:37
Dando18Dando18
4,68241235
answered Dec 11 '18 at 2:37
Dando18Dando18
4,68241235
answered Dec 11 '18 at 2:37
Dando18Dando18
4,68241235
4,68241235
add a comment |
add a comment |
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$begingroup$
$y'(0)=2$ gives you the slope of the line at $(0,1)$. You still need to find the equation of the line (given it's slope and a point lying on it).
$endgroup$
– Anurag A
Dec 11 '18 at 2:28
$begingroup$
I think this might be a case where you're not sure what you're actually looking for. You say you get $2$, which is indeed $y'(0)$. But you're not looking for a number, you're looking for a line.
$endgroup$
– T. Bongers
Dec 11 '18 at 2:30