How do you find the first and second derivatives of #y = (x^2 + 3) / e^-x# using the quotient rule?

Answer 1

# y' = e^x (x^2 + 2x + 3)#
# y'' = e^x (x^2 + 4x + 5)#

Alternatively, you could rewrite the original question as a product and apply the product rule instead of the quotient rule.

STEP 1: Rewrite the original equation #y = (x^2+3)/e^-x = (x^2+3)*e^x#
STEP 2: Use the product rule to find the first derivative #y' = (2x)(e^x) + (x^2+3) * e^(-x) #
STEP 3: Simplify your answer to y' #y' = 2xe^x + x^2e^x + 3e^x = e^x(x^2 + 2x + 3)#
STEP 4: Use the product rule to find the derivative of the derivative, which will give us y'' Recall the product rule is - the derivative of the 1st term, times the 2nd term PLUS the 1st term times the derivative of the 2nd term In this question, our 1st term is: e^x; our 2nd term is: x^2 + 2x + 3 #y'' = e^x * (x^2 + 2x + 3) + e^x * (2x + 2)# # = e^x (x^2 + 2x + 3 + 2x + 2)# # = e^x (x^2 + 4x + 5)#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

To find the first and second derivatives of ( y = \frac{x^2 + 3}{e^{-x}} ) using the quotient rule:

  1. First, let's denote ( u = x^2 + 3 ) and ( v = e^{-x} ).

  2. Then, we'll find the derivatives of ( u ) and ( v ):

    ( u' = 2x )

    ( v' = -e^{-x} )

  3. Next, we'll apply the quotient rule, which states that for functions ( u(x) ) and ( v(x) ), the derivative of their quotient is:

    ( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} )

  4. Now, using the quotient rule, we have:

    ( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{(2x)(e^{-x}) - (x^2 + 3)(-e^{-x})}{(e^{-x})^2} )

  5. Simplifying this expression gives us the first derivative:

    ( \frac{d}{dx}\left(\frac{x^2 + 3}{e^{-x}}\right) = \frac{2xe^{-x} + (x^2 + 3)e^{-x}}{e^{-2x}} )

  6. To find the second derivative, we need to differentiate the first derivative again with respect to ( x ):

    ( \frac{d^2}{dx^2}\left(\frac{x^2 + 3}{e^{-x}}\right) = \frac{d}{dx}\left(\frac{2xe^{-x} + (x^2 + 3)e^{-x}}{e^{-2x}}\right) )

  7. Simplify the expression obtained from the first derivative and then apply the quotient rule again to find the second derivative.

  8. After simplifying the expression obtained for the second derivative, you'll have the result.

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7