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junior college 1 | H2 Maths
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Sonia
Sonia

junior college 1 chevron_right H2 Maths chevron_right Singapore

Hi!! Can someone please help me with implicit differentiation for these questions? I know it’s basic but i can’t even do the first one because i don’t know if i’m just supposed to multiply d/dx inside :( Thank u so much!!

Date Posted: 4 years ago
Views: 278
J
J
4 years ago
First one :

dy/dx means differentiation of y with respect to x, whereby y is a function of x.

i.e y = f(x)


So d/dx (dy/dx) just means to differentiate this with respect to x again

d/dx (dy/dx) = d²y/dx²

This is the second derivative.


Alternative representation :


y = f(x)

dy/dx = d/dx ( f(x) ) or d(f(x))/dx= f'(x)

d/dx (dy/dx) = d/dx ( f'(x) ) or d(f'(x))/dx
= f''(x)
= d²y/dx²
J
J
4 years ago
d/dx does not mean multiplying.

It means differentiating with respect to x.

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Khir Zulkifli
Khir Zulkifli's answer
29 answers (Tutor Details)
1st
Hi I have given a helping hand by solving 2 questions. For implicit differentiation if it’s d/dx(dy/dy) you will just get the second order of differentiation so it’s d^2y/dx^2.
J
J
4 years ago
d/dx (dy/dx) rather than d/dx (dy/dy)
Khir Zulkifli
Khir Zulkifli
4 years ago
Yup a mistake there! Sorry my bad, Thanks!
J
J
4 years ago
No worries, just helping
Eric Nicholas K
Eric Nicholas K
4 years ago
A number of my Sec 4 students think that dy/dx (dy/dx) is no different from d/dx (dy/dx). When they started learning it they did not think that it would actually be different. It’s not that explicitly taught in Sec 4.
J
J
4 years ago
It is important to recognise that d/dx is an operator, not a fraction
J
J
4 years ago
dy/dx means the operator has been applied on y already. Hence dy/dx (dy/dx)
Is simply (dy/dx)²

Whereas d/dx (dy/dx) is an operation on the first derivative. This leads to the second derivative which is represented by d²y/dx² (also can be seen to be applying the operator twice on y itself).

Teachers should have made this clear in O level teaching.