# How would you determine dy/dx for `y=cos^3(t^5-2)` , `x= sin(5t^6)` ?

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### 1 Answer

`y=cos^3(t^5-2)`

`x=sin(5t^6)`

First, take the derivative of y with respect to t. Use the power formula `(u^n)' = n*u^(n-1) *u'` .

`(dy)/(dt) = 3cos^2(t^5-2) *(cos (t^5-2) '`

To determine `cos (t^5-2)` , apply the formula `(cosu)' = -sin u* u'` .

`(dy)/(dt) =3cos^2(t^5-2) *(-sin (t^5-2))*(t^5)'`

For `(t^5)'` , apply the power formula again.

`(dy)/(dt)= 3cos^2(t^5-2)*(-sin(t^5-2))*5t^4`

`(dy)/(dt) = -15t^4 cos^2(t^5-2)sin(t^5-2)`

Next, determine the derivative of x with respect to t. Apply the formula `(sinu)'=cos u * u'` .

`(dx)/(dt) = cos(5t^6)* (5t^6)'`

Use the power formula again.

`(dx)/(dt) = cos (5t^6)*5*6t^5`

`(dx)/(dt) = 30t^5cos (5t^6)`

Then, divide `(dy)/(dt)` by `(dx)/(dt)` to get `(dy)/(dx)` .

`(dy)/(dx) = ((dy)/(dt))/((dx)/(dt)) = (-15t^4 cos^2(t^5-2)sin(t^5-2))/(30t^5cos (5t^6))`

Cancel common factor between numerator and denominator.

`(dy)/(dx) =- (cos^2(t^5-2)sin(t^5-2))/(2tcos(5t^6))`

**Hence, `(dy)/(dx) =- (cos^2(t^5-2)sin(t^5-2))/(2tcos(5t^6))` .**