Question

find the derivative of y in respect to the appropriate variable.
y=sin^-1 3/t^2

Answer 1

\[y = \sin^{-1}(\frac{3}{t^2})\]?

Answer 2

yeah

Answer 3

\[if \ y = f(g(x))\\then \ y'=f'(g(x))*g'(x)\]

Answer 4

\[Let \ f(x) = \sin^{-1}(x) \\Let \ \ g(x) = \frac{3}{t^2}\\y'=(sin^{-1}(x))^{-1} = \ ?\\g'=(\frac{3}{t^2})' = \ ?\]

Answer 5

\[hint: \ \frac{3}{t^2}=3t^{-2}\]

Answer 6

I thought the negative one was only on the sin why did you place it on the x

Answer 7

\[\frac{1}{x} = x^{-1}\]

Answer 8

the one with sin means function inverse, the 3t^(-1) is multiplicative inverse.

Answer 9

ok but then shouldn't the product rule be implied

Answer 10

no, this is the chain rule

Answer 11

\[y = \sin^{-1}(\frac{3}{t^2})\\y'= \frac{1}{\sqrt{1-(\frac{3}{t^2}})^2}*(\frac{3}{t^2})'\\=\frac{1}{\sqrt{1-\frac{9}{t^4}}}*(-2*3t^{-3})=\frac{-6}{t^3\sqrt{1-\frac{9}{t^4}}}\]

Answer 12

\[\frac{1}{a^b}=a^{-b}\\\frac{1}{a^{-b}}=a^b\]

Answer 13

if that's the anwer than how come in the back of the book it says \[-6/t \sqrt{t^4-9}\]

Answer 14

but you got the t^4 under the 9

Answer 15

\[\frac{-6}{t^3\sqrt{1-\frac{9}{t^4}}}=\frac{-6}{t^3\sqrt{\frac{t^4}{t^4}-\frac{9}{t^4}}}=\frac{-6}{t^3\sqrt{\frac{t^4-9}{t^4}}}=\frac{-6}{t^3\frac{\sqrt{t^4-9}}{\sqrt{t^4}}}\\=\frac{-6}{t^3\frac{\sqrt{t^4-9}}{t^2}}=\frac{-6}{t\sqrt{t^4-9}}\]

Answer 16

no it fine....ohhhh ok thanks... its just i knew how to do i just wasn't getting the same exact answer the book gave and plus i kept forgetting the ^2 after I plugged in for u.

Answer 17

:)