Practice Final #5

We already know the derivative is:

$$f^{\mathrm{\prime }}(x)=\frac{2x}{(1+x^2{)}^2}$$
f'(x) = \frac{2x}{(1 + x^2)^2}


Evaluating it at $x = 1$:

$$f^{\mathrm{\prime }}(1)=\frac{2(1)}{(1+1^2{)}^2}=\frac{2}{4}=\frac{1}{2}$$
f'(1) = \frac{2(1)}{(1 + 1^2)^2} = \frac{2}{4} = \frac{1}{2}


So:

$$m=\frac{1}{2}$$
m = \frac{1}{2}




$$f(1)=\frac{1^2}{1+1^2}=\frac{1}{2}$$
f(1) = \frac{1^2}{1 + 1^2} = \frac{1}{2}


Thus, the point is:

$$(1,\frac{1}{2})$$
(1, \frac{1}{2})




The equation of the line is:

$$y=mx+b$$
y = mx + b


Substitute $m = \frac{1}{2}$ and $(1, \frac{1}{2})$:

$$\frac{1}{2}=\frac{1}{2}(1)+b$$
\frac{1}{2} = \frac{1}{2}(1) + b


Simplify:

$$\frac{1}{2}=\frac{1}{2}+b$$
\frac{1}{2} = \frac{1}{2} + b


Solve for $b$:

$$b=0$$
b = 0




The $y$-intercept is:

$$\boxed{{\displaystyle 0}}$$
\boxed{0}


Thus, the correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}0}}$$
\boxed{(A) \, 0}

We use the quotient rule to differentiate:

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


The quotient rule states:

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


where $u(x) = x^2$ and $v(x) = 1 + x^2$.



$$u^{\mathrm{\prime }}(x)=2x,v^{\mathrm{\prime }}(x)=2x$$
u'(x) = 2x, \quad v'(x) = 2x


Now:

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


Simplify:

$$(2x)(1+x^2)-(x^2)(2x)=2x+2x^3-2x^3=2x$$
(2x)(1 + x^2) - (x^2)(2x) = 2x + 2x^3 - 2x^3 = 2x


Thus:

$$f^{\mathrm{\prime }}(x)=\frac{2x}{(1+x^2{)}^2}$$
f'(x) = \frac{2x}{(1 + x^2)^2}




$$f^{\mathrm{\prime }}(1)=\frac{2(1)}{(1+1^2{)}^2}=\frac{2}{(1+1{)}^2}=\frac{2}{4}=\frac{1}{2}$$
f'(1) = \frac{2(1)}{(1 + 1^2)^2} = \frac{2}{(1 + 1)^2} = \frac{2}{4} = \frac{1}{2}




The slope of the tangent line at $x = 1$ is:

$$\boxed{{\displaystyle \frac{1}{2}}}$$
\boxed{\frac{1}{2}}


Thus, the correct answer is:

$$\boxed{{\displaystyle (B)\hspace{0.17em}\frac{1}{2}}}$$
\boxed{(B) \, \frac{1}{2}}

We are given a right-angled triangle with a hypotenuse that is always 5 cm long. The sides are $x$, $y$, and the hypotenuse is $5$. So the Pythagorean theorem gives:

$$x^2+y^2=5^2=25$$
x^2 + y^2 = 5^2 = 25




$$2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$$
2x \frac{dx}{dt} + 2y \frac{dy}{dt} = 0


Simplify:

$$x\frac{dx}{dt}+y\frac{dy}{dt}=0$$
x \frac{dx}{dt} + y \frac{dy}{dt} = 0




$$y\frac{dy}{dt}=-x\frac{dx}{dt}$$
y \frac{dy}{dt} = -x \frac{dx}{dt}


Divide both sides by $x' \neq 0$:

$$\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$$
\frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}


Thus:

$$\frac{y^{\mathrm{\prime }}}{x^{\mathrm{\prime }}}=-\frac{x}{y}$$
\frac{y'}{x'} = -\frac{x}{y}




Using the Pythagorean theorem:

$$4^2+y^2=25\Rightarrow 16+y^2=25\Rightarrow y^2=9\Rightarrow y=3$$
4^2 + y^2 = 25 \quad \Rightarrow \quad 16 + y^2 = 25 \quad \Rightarrow \quad y^2 = 9 \quad \Rightarrow \quad y = 3




$$\frac{y^{\mathrm{\prime }}}{x^{\mathrm{\prime }}}=-\frac{x}{y}=-\frac{4}{3}$$
\frac{y'}{x'} = -\frac{x}{y} = -\frac{4}{3}




The ratio $\frac{y'}{x'}$ when $x = 4$ is:

$$\boxed{{\displaystyle -\frac{4}{3}}}$$
\boxed{-\frac{4}{3}}


Thus, the correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}-\frac{4}{3}}}$$
\boxed{(A) \, -\frac{4}{3}}

We recognize that the integral of $\frac{1}{1 + x^2}$ is the well-known inverse tangent function:

$$\int \frac{1}{1+x^2}\hspace{0.17em}dx=\arctan (x)+C_1$$
\int \frac{1}{1 + x^2} \, dx = \arctan(x) + C_1




The integral of $\sin(x)$ is:

$$\int \sin (x)\hspace{0.17em}dx=-\cos (x)+C_2$$
\int \sin(x) \, dx = -\cos(x) + C_2




Now we combine the two integrals:

$$\int (\frac{1}{1+x^2}+\sin (x))\hspace{0.17em}dx=\arctan (x)-\cos (x)+C$$
\int \left( \frac{1}{1 + x^2} + \sin(x) \right) \, dx = \arctan(x) - \cos(x) + C


where $C = C_1 + C_2$ is the most general constant of integration.



Thus, the most general antiderivative is:

$$\boxed{{\displaystyle \arctan (x)-\cos (x)+C}}$$
\boxed{\arctan(x) - \cos(x) + C}


The correct answer is:

$$\boxed{{\displaystyle (A)\hspace{0.17em}\arctan (x)-\cos (x)+C}}$$
\boxed{(A) \, \arctan(x) - \cos(x) + C}

We are trying to evaluate:

$$\underset{x\to 0^{+}}{\lim }\frac{\sin (x)(e^x-1)}{x^3}$$
\lim_{x \to 0^+} \frac{\sin(x)(e^x - 1)}{x^3}


We can rewrite this as:

$$\underset{x\to 0^{+}}{\lim }(\frac{\sin (x)}{x}\cdot \frac{e^x-1}{x}\cdot \frac{1}{x})$$
\lim_{x \to 0^+} \left( \frac{\sin(x)}{x} \cdot \frac{e^x - 1}{x} \cdot \frac{1}{x} \right)


By splitting it like this, we can analyze each part separately, making it easier to solve.



1. **Evaluate** $\lim_{x \to 0^+} \frac{\sin(x)}{x}$:

$$\underset{x\to 0^{+}}{\lim }\frac{\sin (x)}{x}=1$$
\lim_{x \to 0^+} \frac{\sin(x)}{x} = 1


This is a standard limit that often appears in exams.

[Remember this limit: $$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=1$$
\lim_{x \to 0} \frac{\sin(x)}{x} = 1

They love to test this! It shows that $\sin(x)$ and $x$ are almost identical for small values of $x$.]

2. **Evaluate** $\lim_{x \to 0^+} \frac{e^x - 1}{x}$:

$$\underset{x\to 0^{+}}{\lim }\frac{e^x-1}{x}=1$$
\lim_{x \to 0^+} \frac{e^x - 1}{x} = 1


For small $x$, the function $e^x - 1$ behaves similarly to $x$.

[This limit is another must-know: $$\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$$
\lim_{x \to 0} \frac{e^x - 1}{x} = 1

This shows that for small $x$, the numerator and denominator act like the same function.]

3. **Evaluate** $\lim_{x \to 0^+} \frac{1}{x}$:

$$\underset{x\to 0^{+}}{\lim }\frac{1}{x}=\mathrm{\infty }$$
\lim_{x \to 0^+} \frac{1}{x} = \infty


This tells us that the function grows without bound as $x$ approaches 0 from the positive side.



Using the property that:

$$\underset{x\to 0^{+}}{\lim }(f(x)\cdot g(x)\cdot h(x))=\underset{x\to 0^{+}}{\lim }f(x)\cdot \underset{x\to 0^{+}}{\lim }g(x)\cdot \underset{x\to 0^{+}}{\lim }h(x)$$
\lim_{x \to 0^+} \left( f(x) \cdot g(x) \cdot h(x) \right) = \lim_{x \to 0^+} f(x) \cdot \lim_{x \to 0^+} g(x) \cdot \lim_{x \to 0^+} h(x)


**(only if all limits are finite or one limit goes to infinity without any undefined behavior like $0 \cdot \infty$)**, we get:

$$1\cdot 1\cdot \mathrm{\infty }=\mathrm{\infty }$$
1 \cdot 1 \cdot \infty = \infty


Since the multiplication involves finite values and one $\infty$, the result is:

$$\boxed{{\displaystyle \mathrm{\infty }}}$$
\boxed{\infty}


This approach works here because each limit we evaluated behaves predictably and does not lead to an undefined form like $0 \cdot \infty$. Therefore, we can confidently multiply the individual limits.


In this tough question they are testing your knowledge of the these two limits:
$$\underset{x\to 0}{\lim }\frac{\sin (x)}{x}=1$$
\lim_{x \to 0} \frac{\sin(x)}{x} = 1
and

$$\underset{x\to 0}{\lim }\frac{e^x-1}{x}=1$$
\lim_{x \to 0} \frac{e^x - 1}{x} = 1
. As long as you know those limits then you'll have no problem with this question.


The reciprocal of each limit is also 1:

$$\underset{x\to 0}{\lim }\frac{x}{\sin (x)}=1\text{and}\underset{x\to 0}{\lim }\frac{x}{e^x-1}=1$$
\lim_{x \to 0} \frac{x}{\sin(x)} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{x}{e^x - 1} = 1


These limits show that $\sin(x)$ and $e^x - 1$ behave almost exactly like $x$ when $x$ is close to 0. Knowing these will help you solve many tricky limits that appear in calculus exams!



You can also therefore mix these terms:

$$\underset{x\to 0}{\lim }\frac{\sin (x)}{e^x-1}=1$$
\lim_{x \to 0} \frac{\sin(x)}{e^x - 1} = 1


Both $\sin(x)$ and $e^x - 1$ behave like $x$ when $x$ is close to 0, so this limit simplifies to:

$$\underset{x\to 0}{\lim }\frac{x}{x}=1$$
\lim_{x \to 0} \frac{x}{x} = 1


This is another handy limit to keep in mind for calculus exams!