Simple Differentiation Question

[eutas1]

Hello, I am completely stuck on this super simple differentiation question. It's literally the first question on a past exam. How do you do this without identities (as I have not learnt any)? I tried using the quotient rule but I did not get the same answer...
Please refer to the attached for the question + answer.
Thank you!

 

[pka]

If [imath]f[/imath] is differentiable function if [imath]x[/imath] then [imath]D_x\left(\dfrac{1}{f}\right)=\left(\dfrac{-f'}{f^2}\right)[/imath]

 

[eutas1]

If [imath]f[/imath] is differentiable function if [imath]x[/imath] then [imath]D_x\left(\dfrac{1}{f}\right)=\left(\dfrac{-f'}{f^2}\right)[/imath]

wait i just realised what i did wrong - i made the derivative of 1 equal to 1 instead of 0.... so stupid! thank you for your time

 

[Steven G]

wait i just realised what i did wrong - i made the derivative of 1 equal to 1 instead of 0.... so stupid! thank you for your time

You really should not use the quotient rule if either the numerator and/or the denominator is a constant.
(1/f)' = (f^-1)' = -1(f^-2)f' = -f'/f²

 

[eutas1]

You really should not use the quotient rule if either the numerator and/or the denominator is a constant.
(1/f)' = (f^-1)' = -1(f^-2)f' = -f'/f²

Ah, I see!! I should have realised that. Thank you!

 

[Harry_the_cat]

I recognise this from the QCAA Methods paper from last year.

It is best to rewrite it as \(\displaystyle y= (sin x)^{-1}\) and then use the chain rule to get \(\displaystyle -1.(sin x)^{-2}.cos x =\frac{-cos x}{(sin x)^2}\)

But the quotient rule will still work and give the same answer (although a bit more work is involved). You must have made a simple error when using the quotient rule.

Using the quotient rule:
Let \(\displaystyle u=1\) and \(\displaystyle v= sin x\). So, \(\displaystyle u'=0\) and \(\displaystyle v' = cos x\)

So \(\displaystyle y' =\frac{u'.v - u.v'}{v^2} = \frac{0.sin x - 1.cos x}{(sin x)^2}\) which is the same answer.