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junior college 1 | H2 Maths
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need help with this qn, pls explain too
Date Posted: 1 year ago
Views: 338
Some comparisons:
d/dx (x²) = 2x
d/dx (y²) = 2y times d/dx (y) = 2x dy/dx (chain rule continues to apply because y is a function of x, albeit an unknown one in this case)
[If you still hold on to a Sec 4 textbook, you would have seen the section on Chain rule where they used the letter u and the notation dy/dx = dy/du du/dx; it's the same idea here]
d/dx (x + 1)³ = 3 (x + 1)² d/dx (x + 1) = 3 (x + 1)²
d/dx (y + 1)³ = 3 (y + 1)² d/dx (y + 1) = 3 (y + 1)² (dy/dx + 0)
d/dx (x²) = 2x
d/dx (y²) = 2y times d/dx (y) = 2x dy/dx (chain rule continues to apply because y is a function of x, albeit an unknown one in this case)
[If you still hold on to a Sec 4 textbook, you would have seen the section on Chain rule where they used the letter u and the notation dy/dx = dy/du du/dx; it's the same idea here]
d/dx (x + 1)³ = 3 (x + 1)² d/dx (x + 1) = 3 (x + 1)²
d/dx (y + 1)³ = 3 (y + 1)² d/dx (y + 1) = 3 (y + 1)² (dy/dx + 0)
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done
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We continue to perform differentiation as usual using regular rules, but this time we do not attempt to make y the subject as we perform the differentiation.
Whenever you perform differentiation last year, in Sec 4,
y = x^2
dy/dx = 2x
What we are actually doing is
y = x^2
d/dx (y) = d/dx (x^2)
The process of writing dy/dx actually has this intermediary step of taking “d/dx” on both sides of the equation.
This is actually what happens for our question - we take d/dx on both sides, and whenever we see a “d/dx (y)”, it becomes dy/dx.
Whenever you perform differentiation last year, in Sec 4,
y = x^2
dy/dx = 2x
What we are actually doing is
y = x^2
d/dx (y) = d/dx (x^2)
The process of writing dy/dx actually has this intermediary step of taking “d/dx” on both sides of the equation.
This is actually what happens for our question - we take d/dx on both sides, and whenever we see a “d/dx (y)”, it becomes dy/dx.
Date Posted:
1 year ago
i dont really understand whats going on before the "now we....." tho
sorry for late reply i didnt get any notifications from the app
sorry for late reply i didnt get any notifications from the app
ohhh i get it now
also, when we differentiate 3y, do we write it as 3(dy/dx) or just dy/dx?
since d/dx(y) = dy/dx, but now theres a 3 infront and im not sure how to deal with it...
if uts d/dx(3x) its just 3 but i dont think d/dx(3y)=3
also, when we differentiate 3y, do we write it as 3(dy/dx) or just dy/dx?
since d/dx(y) = dy/dx, but now theres a 3 infront and im not sure how to deal with it...
if uts d/dx(3x) its just 3 but i dont think d/dx(3y)=3
d/dx (3y)
= 3 times d/dx (y)
= 3 times dy/dx
= 3 dy/dx
dy/dx is in fact an expression for d/dx (y)
In general, whenever you do the differential of something which is y, we do everything the same as though it is an x, but we add an extra dy/dx behind as part of the chain rule
Eg
d/dx (10x^2) = 20x
d/dx (10y^2) = 20y times dy/dx
NOTE: d/dy (10y^2) = 20y without the dy/dx
= 3 times d/dx (y)
= 3 times dy/dx
= 3 dy/dx
dy/dx is in fact an expression for d/dx (y)
In general, whenever you do the differential of something which is y, we do everything the same as though it is an x, but we add an extra dy/dx behind as part of the chain rule
Eg
d/dx (10x^2) = 20x
d/dx (10y^2) = 20y times dy/dx
NOTE: d/dy (10y^2) = 20y without the dy/dx
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