# a. Compute \frac{d}{dx} \frac{x^2 - 1}{(x^2 + 3)^3} b. Find lim_{x \rightarrow \infty} \sqrt{x^2...

## Question:

a. Compute {eq}\frac{d}{dx} \frac{x^2 - 1}{(x^2 + 3)^3}{/eq}

b. Find {eq}lim_{x \rightarrow \infty} \sqrt{x^2 + x} - x {/eq}

c. Find {eq}lim_{s \rightarrow \infty} \frac{\sqrt{s + 3} + \sqrt{s - 4}}{\sqrt{s + 1} + \sqrt{s + 5}} {/eq}

d. Compute {eq}\frac{d}{dx} \frac{x + 1}{x^2 e^x} {/eq}

e. Compute {eq}\frac{d}{d \phi} \frac{1 + tan( \phi)}{sec( \phi) + 3}{/eq}.

## Quotient and Limit Rule:

If there is quotient expression containing algebraic, exponential and trig-functions such as {eq}\displaystyle \frac{P(u)}{Q(u)} {/eq} and {eq}P'(u), \ Q'(u) {/eq} are first derivative functions with respect to the varaible {eq}u {/eq}, then we'll use the formula for the derivative shown below:

• {eq}\displaystyle \frac{\mathrm{d} }{\mathrm{d} u}\left (\frac{P(u)}{Q(u)} \right )=\frac{Q(u)P'(u)-P(u)Q'(u)}{(Q(u))^2} {/eq}

For the indeterminate form {eq}\displaystyle \frac{\infty }{\infty } {/eq} of limit, we'll identify the highest power of denominator (Suppose it is {eq}f(x) {/eq}) and then divide both the numerator and denominator by {eq}f(x) {/eq} to simplify the limit.

a.

The given expression for the first derivative function is:

{eq}\displaystyle \frac{d}{dx} \frac{x^2 - 1}{(x^2 + 3)^3}=? {/eq}

Applying the quotient rule of derivatives in the above expression, we get:

{eq}\begin{align*} \displaystyle \frac{d}{dx} \left (\frac{x^2 - 1}{(x^2 + 3)^3} \right )&=\frac{(x^2 + 3)^3\frac{\mathrm{d} (x^2 - 1)}{\mathrm{d} x}-(x^2 - 1)\frac{\mathrm{d} (x^2 + 3)^3}{\mathrm{d} x}}{((x^2 + 3)^3)^2}\\ &=\displaystyle \frac{(x^2 + 3)^3(2x - 0)-(x^2 - 1)3(x^2 + 3)^2\frac{\mathrm{d} (x^2 + 3)}{\mathrm{d} x}}{(x^2 + 3)^6}\\ &=\displaystyle \frac{2x(x^2 + 3)^3-3(x^2 - 1)(x^2 + 3)^2(2x +0)}{(x^2 + 3)^6}\\ &=\displaystyle \frac{2x(x^2 + 3)^3-6x(x^2 - 1)(x^2 + 3)^2}{(x^2 + 3)^6}\\ &=\displaystyle \frac{2x(x^2 + 3)^2(x^2 + 3-3(x^2 - 1))}{(x^2 + 3)^6}\\ &=\displaystyle \frac{2x(x^2 + 3-3x^2 +3)}{(x^2 + 3)^4}\\ &=\displaystyle \frac{2x(6-2x^2)}{(x^2 + 3)^4}\\ \end{align*} {/eq}

b.

The limit expression is:

{eq}L=\lim_{x \rightarrow \infty} \sqrt{x^2 + x} - x {/eq}

The value of the above expression at {eq}x=\infty {/eq} is equal to {eq}\infty-\infty {/eq}.

We'll use the rationalization method to simplify the indeterminate form of the given limit expression.

{eq}\begin{align*} \displaystyle L&=\lim_{x \rightarrow \infty} \frac{\sqrt{x^2 + x} - x}{1} \times \frac{\sqrt{x^2 + x}+ x}{\sqrt{x^2 + x} + x}\\ &=\lim_{x \rightarrow \infty} \frac{(\sqrt{x^2 + x} - x)(\sqrt{x^2 + x} + x)}{\sqrt{x^2 + x} + x}\\ &=\lim_{x \rightarrow \infty} \frac{(\sqrt{x^2 + x})^2 - x^2}{\sqrt{x^2 + x} + x}&\because m^2-n^2=(m-n)(m+n)\\ &=\lim_{x \rightarrow \infty} \frac{x^2 + x - x^2}{\sqrt{x^2 + x} + x}\\ &=\lim_{x \rightarrow \infty} \frac{x }{\sqrt{x^2 + x} + x}\\ &=\lim_{x \rightarrow \infty} \frac{x }{\sqrt{x^2 + x} + x}\\ \end{align*} {/eq}

Further, simplify the above expression.

{eq}\begin{align*} \displaystyle L&=\displaystyle \lim_{x \rightarrow \infty} \frac{x }{\sqrt{x^2\left (1+\frac{ x }{x^2} \right )} + x}\\ &=\displaystyle \lim_{x \rightarrow \infty} \frac{x }{\sqrt{x^2}\sqrt{1+\frac{ 1 }{x}} + x}&\because \sqrt{ab}=\sqrt{a}\cdot \sqrt{b}\\ &=\displaystyle \lim_{x \rightarrow \infty} \frac{x }{x\sqrt{1+\frac{ 1 }{x}} + x}\\ &=\displaystyle \lim_{x \rightarrow \infty} \frac{x }{x\left (\sqrt{1+\frac{ 1 }{x}} + 1 \right )}\\ &=\displaystyle \lim_{x \rightarrow \infty} \frac{1 }{\sqrt{1+\frac{ 1 }{x}} + 1 }\\ &= \displaystyle \frac{1 }{\sqrt{1+\frac{ 1 }{\infty}} + 1 }\\ &= \displaystyle \frac{1 }{\sqrt{1+0} + 1 }&\because \frac{ 1 }{\infty}=0\\ &=\displaystyle \frac{1 }{2 } \end{align*} {/eq}

c.

The quotient limit expression is:

{eq}L=\displaystyle \lim_{s \rightarrow \infty} \frac{\sqrt{s + 3} + \sqrt{s - 4}}{\sqrt{s + 1} + \sqrt{s + 5}} {/eq}

The value of above expression at {eq}\infty {/eq} is equal to {eq}\displaystyle \frac{ \infty}{ \infty} {/eq} so we'll divide both numerator and denominator by the highest denominator power.

{eq}\begin{align*} L&=\displaystyle \lim_{s \rightarrow \infty} \frac{\displaystyle \frac{\sqrt{s + 3} + \sqrt{s - 4}}{\sqrt{s + 1}}}{\displaystyle \frac{\sqrt{s + 1} + \sqrt{s + 5}}{\sqrt{s + 1}}}\\ &=\displaystyle \lim_{s \rightarrow \infty} \frac{\displaystyle \frac{\sqrt{s + 3} }{\sqrt{s + 1}}+ \frac{\sqrt{s - 4}}{\sqrt{s + 1}}}{\displaystyle 1+\frac{\sqrt{s + 5}}{\sqrt{s + 1}}}\\ &=\displaystyle \frac{\displaystyle \lim_{s \rightarrow \infty}\left (\frac{\sqrt{s + 3} }{\sqrt{s + 1}}+ \frac{\sqrt{s - 4}}{\sqrt{s + 1}} \right )}{\displaystyle\lim_{s \rightarrow \infty} \left (1+\frac{\sqrt{s + 5}}{\sqrt{s + 1}} \right )}\\ &=\displaystyle \frac{\displaystyle \lim_{s \rightarrow \infty}\left (\frac{\sqrt{s}\sqrt{1 + \frac{3}{s}} }{\sqrt{s}\sqrt{1 + \frac{1}{s}}}+ \frac{\sqrt{s}\sqrt{1 - \frac{4}{s}}}{\sqrt{s}\sqrt{1+ \frac{1}{s}}} \right )}{\displaystyle\lim_{s \rightarrow \infty} \left (1+\frac{\sqrt{s}\sqrt{1 + \frac{5}{s}}}{\sqrt{s}\sqrt{1 + \frac{1}{s}}} \right )}\\ \end{align*} {/eq}

Further, simplify the above expression.

{eq}\begin{align*} L&=\displaystyle \frac{\displaystyle \lim_{s \rightarrow \infty}\left (\frac{\sqrt{1 + \frac{3}{s}} }{\sqrt{1 + \frac{1}{s}}}+ \frac{\sqrt{1 - \frac{4}{s}}}{\sqrt{1+ \frac{1}{s}}} \right )}{\displaystyle\lim_{s \rightarrow \infty} \left (1+\frac{\sqrt{1 + \frac{5}{s}}}{\sqrt{1 + \frac{1}{s}}} \right )}\\ &=\displaystyle \frac{\displaystyle \left (\frac{\sqrt{1 + \frac{3}{\infty}} }{\sqrt{1 + \frac{1}{\infty}}}+ \frac{\sqrt{1 - \frac{4}{\infty}}}{\sqrt{1+ \frac{1}{\infty}}} \right )}{\displaystyle \left (1+\frac{\sqrt{1 + \frac{5}{\infty}}}{\sqrt{1 + \frac{1}{\infty}}} \right )}\\ &=\displaystyle \frac{\displaystyle \left (\frac{\sqrt{1 + 0} }{\sqrt{1 + 0}}+ \frac{\sqrt{1 - 0}}{\sqrt{1+ 0}} \right )}{\displaystyle \left (1+\frac{\sqrt{1 + 0}}{\sqrt{1 + 0}} \right )}&\because \frac{1}{\infty}=0\\ &=\displaystyle \frac{1+ 1}{1+1}\\ &=\displaystyle \frac{2}{2}\\ &=1 \end{align*} {/eq}

d.

The derivative expression is:

{eq}\displaystyle \frac{d}{dx} \left ( \frac{x + 1}{x^2 e^x} \right) {/eq}

By the quotient rule of derivatives, we'll simplify the above quotient expression.

{eq}\begin{align*} \displaystyle \frac{d}{dx} \left ( \frac{x + 1}{x^2 e^x} \right)&=\frac{(x^2 e^x)\frac{\mathrm{d}(x + 1) }{\mathrm{d} x}+(x + 1)\frac{\mathrm{d} (x^2 e^x)}{\mathrm{d} x}}{(x^2 e^x)^2} \\ &=\frac{(x^2 e^x)(1 + 0)+(x + 1)\left (x^2 \frac{\mathrm{d} (e^x)}{\mathrm{d} x} +e^x\frac{\mathrm{d} (x^2 )}{\mathrm{d} x} \right )}{(x^2 e^x)^2} \\ &=\frac{x^2 e^x+(x + 1)\left (x^2e^x + e^x(2x ) \right )}{(x^2 e^x)^2} \\ &=\frac{x^2 e^x+(x + 1)\left (x^2e^x + 2xe^x \right )}{(x^2 e^x)^2} \\ \end{align*} {/eq}

e.

The quotient expression for derivative is:

{eq}\displaystyle \frac{d}{d \phi}\left ( \frac{1 + \tan( \phi)}{\sec( \phi) + 3}\right ) {/eq}

According the quotient rule of derivatives, the simplification of the above expression is:

{eq}\begin{align*} \displaystyle \frac{d}{d \phi}\left ( \frac{1 + \tan( \phi)}{\sec( \phi) + 3}\right )&=\frac{(\sec( \phi) + 3)\frac{\mathrm{d}(1 + \tan( \phi)) }{\mathrm{d} x}+(1 + \tan( \phi))\frac{\mathrm{d} (\sec( \phi) + 3)}{\mathrm{d} x}}{(\sec( \phi) + 3)^2}\\ &=\frac{(\sec( \phi) + 3)(0 + \sec^2( \phi))+(1 + \tan( \phi))(\sec( \phi)\tan( \phi) + 0)}{(\sec( \phi) + 3)^2}\\ &=\frac{\sec^2( \phi)(\sec( \phi) + 3)+\sec( \phi)\tan( \phi) (1 + \tan( \phi))}{(\sec( \phi) + 3)^2}\\ \end{align*} {/eq}