Question

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

Answer 1

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

Answer 2

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

Answer 3

@completeidiot I wrote the exact same thing... It was just a different form.

@satellite73 Could you explain that?

Answer 4

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

Answer 5

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

Answer 6

@satellite73

Answer 7

ok lets use an example

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

now you can cancel a bunch of these

Answer 15

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

Answer 16

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

Answer 17

Thank you both so much! @satellite73 @genius12