Calculating the differential and relating it to the directional derivative
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I need to calculate the differential of the following function, but I am struggling to make sense of the relationship between the directional derivative and differential:
$f(x,y) = ln(x^2+2y+1) + int_0^x cos(t^2) dt$
I know the partials are:
$frac{df}{dx} = frac{2x}{x^2+2y+1} + cos(x^2)$
$frac{df}{dy} = frac{2}{x^2+2y+1}$
I thought that the differential $L_{x_0}$ would be $[frac{df}{dx} frac{df}{dy}]'$ (that is the transpose of the gradient). However, the answer given is:
$L_{x_0}(v)= L_{x_0} . v = frac{df}{dx} v_1 + frac{df}{dy}v_2$
I am a bit confused about this. Is the differential $L_{x_0}(v) = L_{x_0} . v$ (which I thought was the directional derivative) or just $L_{x_0}$? Do I need to define the differential in terms of a direction?
Thanks for your help.
multivariable-calculus
asked Dec 6 '18 at 12:20
ChristianChristian
367
$endgroup$
add a comment |
$begingroup$
I need to calculate the differential of the following function, but I am struggling to make sense of the relationship between the directional derivative and differential:
$f(x,y) = ln(x^2+2y+1) + int_0^x cos(t^2) dt$
I know the partials are:
$frac{df}{dx} = frac{2x}{x^2+2y+1} + cos(x^2)$
$frac{df}{dy} = frac{2}{x^2+2y+1}$
I thought that the differential $L_{x_0}$ would be $[frac{df}{dx} frac{df}{dy}]'$ (that is the transpose of the gradient). However, the answer given is:
$L_{x_0}(v)= L_{x_0} . v = frac{df}{dx} v_1 + frac{df}{dy}v_2$
I am a bit confused about this. Is the differential $L_{x_0}(v) = L_{x_0} . v$ (which I thought was the directional derivative) or just $L_{x_0}$? Do I need to define the differential in terms of a direction?
Thanks for your help.
multivariable-calculus
asked Dec 6 '18 at 12:20
ChristianChristian
367
$endgroup$
add a comment |
$begingroup$
I need to calculate the differential of the following function, but I am struggling to make sense of the relationship between the directional derivative and differential:
$f(x,y) = ln(x^2+2y+1) + int_0^x cos(t^2) dt$
I know the partials are:
$frac{df}{dx} = frac{2x}{x^2+2y+1} + cos(x^2)$
$frac{df}{dy} = frac{2}{x^2+2y+1}$
I thought that the differential $L_{x_0}$ would be $[frac{df}{dx} frac{df}{dy}]'$ (that is the transpose of the gradient). However, the answer given is:
$L_{x_0}(v)= L_{x_0} . v = frac{df}{dx} v_1 + frac{df}{dy}v_2$
I am a bit confused about this. Is the differential $L_{x_0}(v) = L_{x_0} . v$ (which I thought was the directional derivative) or just $L_{x_0}$? Do I need to define the differential in terms of a direction?
Thanks for your help.
multivariable-calculus
asked Dec 6 '18 at 12:20
ChristianChristian
367
$endgroup$
I need to calculate the differential of the following function, but I am struggling to make sense of the relationship between the directional derivative and differential:
$f(x,y) = ln(x^2+2y+1) + int_0^x cos(t^2) dt$
I know the partials are:
$frac{df}{dx} = frac{2x}{x^2+2y+1} + cos(x^2)$
$frac{df}{dy} = frac{2}{x^2+2y+1}$
I thought that the differential $L_{x_0}$ would be $[frac{df}{dx} frac{df}{dy}]'$ (that is the transpose of the gradient). However, the answer given is:
$L_{x_0}(v)= L_{x_0} . v = frac{df}{dx} v_1 + frac{df}{dy}v_2$
I am a bit confused about this. Is the differential $L_{x_0}(v) = L_{x_0} . v$ (which I thought was the directional derivative) or just $L_{x_0}$? Do I need to define the differential in terms of a direction?
Thanks for your help.
multivariable-calculus
multivariable-calculus
asked Dec 6 '18 at 12:20
ChristianChristian
367
asked Dec 6 '18 at 12:20
ChristianChristian
367
asked Dec 6 '18 at 12:20
ChristianChristian
367
asked Dec 6 '18 at 12:20
ChristianChristian
367
asked Dec 6 '18 at 12:20
ChristianChristian
367
367
add a comment |
add a comment |
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