Question

Find the first and second derivatives of the function.\\ y = \frac{(x+3)(x^2 - 3x + 9)}{x^3}\\ The first derivative of the function is $\frac{dy}{dx} = -\frac{81}{4x}$.\\The second derivative of the function is $\frac{d^2y}{dx^2} = \square$.

          Find the first and second derivatives of the function.\\
y = \frac{(x+3)(x^2 - 3x + 9)}{x^3}\\
The first derivative of the function is $\frac{dy}{dx} = -\frac{81}{4x}$.\\The second derivative of the function is $\frac{d^2y}{dx^2} = \square$.

Find the first and second derivatives of the function.

y = ((x+3)(x^2 - 3x + 9))/(x^3)

The first derivative of the function is (dy)/(dx) = -(81)/(4x).
The second derivative of the function is (d^2y)/(dx^2) = □.

Added by Mario B.

Calculus: Early Transcendentals

James Stewart 8th Edition

Instant Answer

Solved by Expert Manisha Sarker

06/04/2021

Step 1

The given function is +32-3+9. This can be simplified to +38. Show more…

Show all steps

Thanks for your feedback!

Find the first and second derivatives of the function +32-3+9 dy The first derivative of the function is dx 81 a2y The second derivative of the function is dx