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How do you do chain rule? - Answers
The chain rule: given two functions of x, u & v, and let their respective derivatives with respect to x denote by u' and v', the derivative with respect to x of (u*v) is: v*u' + u*v' For division of two functions (u/v), derivative of (u/v) = [v*u' - u*v']/(v2) If you cannot remember if you've got it right, try this simple check: take f(x) = x3 but let u = x and v = x2, so u*v = x3. Derivative of x3 = 3*x2. Now try it with the chain rule: d(x*x2) = x2*1 + x*(2*x2)= 3*x2 I think the product rule is easier to remember, but sometimes cannot remember the quotient rule, so I'll try a simple one to make sure I got it right. Take u = x3 and let v = x, so u/v = x2, which the derivative = 2*x So for [v*u' - u*v']/(v2) we have: [x*3*x2 - x3*1]/(x2) = [2*x3]/(x2) = 2*x, so I did it correctly. I'll show one example of the product rule for more complex function, take sin(x)*ex --> u = sin(x), v=ex, so u' = cos(x) & v' = ex d(u*v) = ex * sin(x) + ex * cos(x) = (sin(x) + cos(x)) * ex
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How do you do chain rule? - Answers
The chain rule: given two functions of x, u & v, and let their respective derivatives with respect to x denote by u' and v', the derivative with respect to x of (u*v) is: v*u' + u*v' For division of two functions (u/v), derivative of (u/v) = [v*u' - u*v']/(v2) If you cannot remember if you've got it right, try this simple check: take f(x) = x3 but let u = x and v = x2, so u*v = x3. Derivative of x3 = 3*x2. Now try it with the chain rule: d(x*x2) = x2*1 + x*(2*x2)= 3*x2 I think the product rule is easier to remember, but sometimes cannot remember the quotient rule, so I'll try a simple one to make sure I got it right. Take u = x3 and let v = x, so u/v = x2, which the derivative = 2*x So for [v*u' - u*v']/(v2) we have: [x*3*x2 - x3*1]/(x2) = [2*x3]/(x2) = 2*x, so I did it correctly. I'll show one example of the product rule for more complex function, take sin(x)*ex --> u = sin(x), v=ex, so u' = cos(x) & v' = ex d(u*v) = ex * sin(x) + ex * cos(x) = (sin(x) + cos(x)) * ex
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How do you do chain rule? - Answers
The chain rule: given two functions of x, u & v, and let their respective derivatives with respect to x denote by u' and v', the derivative with respect to x of (u*v) is: v*u' + u*v' For division of two functions (u/v), derivative of (u/v) = [v*u' - u*v']/(v2) If you cannot remember if you've got it right, try this simple check: take f(x) = x3 but let u = x and v = x2, so u*v = x3. Derivative of x3 = 3*x2. Now try it with the chain rule: d(x*x2) = x2*1 + x*(2*x2)= 3*x2 I think the product rule is easier to remember, but sometimes cannot remember the quotient rule, so I'll try a simple one to make sure I got it right. Take u = x3 and let v = x, so u/v = x2, which the derivative = 2*x So for [v*u' - u*v']/(v2) we have: [x*3*x2 - x3*1]/(x2) = [2*x3]/(x2) = 2*x, so I did it correctly. I'll show one example of the product rule for more complex function, take sin(x)*ex --> u = sin(x), v=ex, so u' = cos(x) & v' = ex d(u*v) = ex * sin(x) + ex * cos(x) = (sin(x) + cos(x)) * ex
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- og:descriptionThe chain rule: given two functions of x, u & v, and let their respective derivatives with respect to x denote by u' and v', the derivative with respect to x of (u*v) is: v*u' + u*v' For division of two functions (u/v), derivative of (u/v) = [v*u' - u*v']/(v2) If you cannot remember if you've got it right, try this simple check: take f(x) = x3 but let u = x and v = x2, so u*v = x3. Derivative of x3 = 3*x2. Now try it with the chain rule: d(x*x2) = x2*1 + x*(2*x2)= 3*x2 I think the product rule is easier to remember, but sometimes cannot remember the quotient rule, so I'll try a simple one to make sure I got it right. Take u = x3 and let v = x, so u/v = x2, which the derivative = 2*x So for [v*u' - u*v']/(v2) we have: [x*3*x2 - x3*1]/(x2) = [2*x3]/(x2) = 2*x, so I did it correctly. I'll show one example of the product rule for more complex function, take sin(x)*ex --> u = sin(x), v=ex, so u' = cos(x) & v' = ex d(u*v) = ex * sin(x) + ex * cos(x) = (sin(x) + cos(x)) * ex
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