Can anyone help me find the derivative without using the quotient rule of the following equation?

\[2 \div ((2t ^{2} - 4\sin(t))^{5})\]

\[\left(\frac{1}{f}\right)'=-\frac{f'}{f^2}\]

@completeidiot I wrote the exact same thing... It was just a different form. @satellite73 Could you explain that?

it is a short way to say what the derivative of the reciprocal of a function is

I don't understand how that helps? Sorry ><

@satellite73

ok lets use an example

find the derivative of \(\frac{1}{x^2+2x}\)

here we have \(\frac{1}{f}\) where \(f(x)=x^2+2x\) so the derivative is \[-\frac{2x+2}{(x^2+2x)^2}\]

i was using the formula i wrote above \[\left(\frac{1}{f}\right)'=-\frac{f'}{f^2}\]

here is another example \[\frac{5}{\sin(x)+x^2}\] the derivative will be \[-\frac{5(\cos(x)+2x)}{(\sin(x)+x^2)^2}\]

your example is somewhat more complicated here \(f(t)=(2t^2-4\sin(t))^5\) so \(f'(t)=5(2t^2-4\sin(t))^4(4t-4\cos(t))\)

you get as your derivative \[-\frac{10(2t^2-4\sin(t))^4(4t-4\cos(t))}{(2t^2-4\sin(t))^{10}}\]

now you can cancel a bunch of these

Why don't you write the denomination with a rational exponent and then use the Power Rule, Constant Multiple Rule along with Chain Rule to differentiate? What I mean is that you could re-write it the problem as this:\[f(t)=\frac{ 2 }{ [2t^2-4\sin(t)]^5 }=2*[2t^2-4\sin(t)]^{-5}\]Now you simply differentiate the way you would with Power Rule, Chain Rule, and Constant Multiple Rule the way you normally would. You could alternatively use the Reciprocal Rule as satellite73 has already suggested. @alaskamath

\[-\frac{10(4t-4\cos(t))}{(2t^2-4\sin(t))^{6}}\]

Thank you both so much! @satellite73 @genius12

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