How do you differentiate #f(x)=(x^2+e^x)/(x-1)^2# using the quotient rule?

Question

Paisley Johnson · Accepted Answer

#d/dx (x^2+e^x)/(x-1)^2=((x-1)^2*(2x+e^x)-2(x-1)*(x^2+e^x))/((x-1)^2)#

The quotient rule states that #d/dxf(x)/(g(x))=(g(x)*f'(x)-f(x)*g'(x))/(g(x))^2#

So in this case,

#d/dx (x^2+e^x)/(x-1)^2=((x-1)^2*(2x+e^x)-2(x-1)*(x^2+e^x))/((x-1)^2)#

Isabella Ackerman · Answer

undefined To differentiate $ f(x) = \frac{{x^2 + e^x}}{{(x - 1)^2}} $ using the quotient rule:

1. Identify $ u(x) = x^2 + e^x $ and $ v(x) = (x - 1)^2 $.
2. Apply the quotient rule:

$$ f'(x) = \frac{{v(x) \cdot u'(x) - u(x) \cdot v'(x)}}{{[v(x)]^2}} $$

3. Compute the derivatives:
   - $ u'(x) = 2x + e^x $
   - $ v'(x) = 2(x - 1) $

4. Substitute the derivatives and functions into the quotient rule formula.
5. Simplify the expression.

Paisley Johnson · Answer

undefined To differentiate the function $f(x) = \frac{x^2 + e^x}{(x - 1)^2}$ using the quotient rule, we follow these steps:

1. Let $u(x) = x^2 + e^x$ and $v(x) = (x - 1)^2$.
2. Apply the quotient rule formula, which states that the derivative of a quotient of two functions $u(x)$ and $v(x)$ is given by:

$$f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$$

3. Calculate $u'(x)$ and $v'(x)$.
   - $u'(x) = \frac{d}{dx}(x^2 + e^x) = 2x + e^x$
   - $v'(x) = \frac{d}{dx}((x - 1)^2) = 2(x - 1)$

4. Substitute the derivatives and original functions into the quotient rule formula:

$$f'(x) = \frac{(2x + e^x)(x - 1)^2 - (x^2 + e^x)(2(x - 1))}{((x - 1)^2)^2}$$

5. Simplify the expression:

$$f'(x) = \frac{(2x + e^x)(x^2 - 2x + 1) - 2(x^2 + e^x)(x - 1)}{(x - 1)^4}$$

6. Expand and combine like terms:

$$f'(x) = \frac{2x^3 - 4x^2 + 2x + xe^x - 2e^x - 2x^3 + 2x^2 + 2xe^x - 2e^x}{(x - 1)^4}$$

$$f'(x) = \frac{-2x^2 + 4x - 2e^x}{(x - 1)^4}$$

Thus, the derivative of $f(x) = \frac{x^2 + e^x}{(x - 1)^2}$ using the quotient rule is $f'(x) = \frac{-2x^2 + 4x - 2e^x}{(x - 1)^4}$.

Samantha Adkison · Answer

undefined To differentiate the function $ f(x) = \frac{x^2 + e^x}{(x - 1)^2} $ using the quotient rule, we apply the formula:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} $$

Let $ u(x) = x^2 + e^x $ and $ v(x) = (x - 1)^2 $. Then, we find the derivatives:

$$ u'(x) = 2x + e^x $$
$$ v'(x) = 2(x - 1) $$

Now, we substitute these values into the quotient rule formula:

$$ \frac{d}{dx} \left( \frac{x^2 + e^x}{(x - 1)^2} \right) = \frac{(2x + e^x)(x - 1)^2 - (x^2 + e^x)(2(x - 1))}{[(x - 1)^2]^2} $$

$$ = \frac{(2x + e^x)(x^2 - 2x + 1) - 2(x^2 + e^x)(x - 1)}{(x - 1)^4} $$

$$ = \frac{2x^3 - 4x^2 + 2x + xe^x - 2e^x - 2x^3 + 2x^2 + 2xe^x - 2e^x}{(x - 1)^4} $$

$$ = \frac{-2x^2 + 4x - 2e^x}{(x - 1)^4} $$

Thus, the derivative of the function $ f(x) = \frac{x^2 + e^x}{(x - 1)^2} $ using the quotient rule is:

$$ f'(x) = \frac{-2x^2 + 4x - 2e^x}{(x - 1)^4} $$