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Problem 970

Find the derivative of: $$f(x)=\left(x^{2}+1\right) /(1-3 x)$$.

Expert verified
The short answer for the derivative of $$f(x) = \frac{x^2 + 1}{1 - 3x}$$ is $$f'(x) = \frac{-3x^2 + 2x + 3}{(1 - 3x)^2}$$.
See the step by step solution

Step 1: Identify the numerator and denominator functions

We have a function in the form of $$\frac{u(x)}{v(x)}$$, where $$u(x) = x^2 + 1$$ and $$v(x) = 1 - 3x$$.

Step 2: Calculate the derivatives of the numerator and denominator functions

Now we will find the derivatives of the two functions - $$u'(x)$$ and $$v'(x)$$. For $$u(x) = x^2 + 1$$, use the Power Rule: $u'(x) = \frac{d}{dx}(x^2) + \frac{d}{dx}(1) = 2x + 0 = 2x$ For $$v(x) = 1 - 3x$$, again, use the Power Rule: $v'(x) = \frac{d}{dx}(1) - \frac{d}{dx}(3x) = 0 - 3 = -3$

Step 3: Apply the Quotient Rule

Now that we have found $$u'(x)$$ and $$v'(x)$$, we can apply the Quotient Rule to find the derivative of $$f(x)$$. The Quotient Rule states: $\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$ Plug in the expressions we found in Steps 1 and 2: $\frac{d}{dx}\left(\frac{x^2 + 1}{1 - 3x}\right) = \frac{(2x)(1 - 3x) - (x^2 + 1)(-3)}{(1 - 3x)^2}$

Step 4: Simplify the expression

Now, simplify the expression obtained in Step 3: $\frac{d}{dx}\left(\frac{x^2 + 1}{1 - 3x}\right) = \frac{2x(1 - 3x) + 3(x^2 + 1)}{(1 - 3x)^2}$ Expand the numerator and combine like terms: $\frac{d}{dx}\left(\frac{x^2 + 1}{1 - 3x}\right) = \frac{2x - 6x^2 + 3x^2 + 3}{(1 - 3x)^2}$ $\frac{d}{dx}\left(\frac{x^2 + 1}{1 - 3x}\right) = \frac{-3x^2 + 2x + 3}{(1 - 3x)^2}$ So the derivative of $$f(x) = \frac{x^2 + 1}{1 - 3x}$$ is: $f'(x) = \frac{-3x^2 + 2x + 3}{(1 - 3x)^2}$

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