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Tangent Lines


  1. The slope of the tangent line is simply the derivative evaluated at a given point (in this case $x=3$). This means that all we have to find is $f'(3)$. To do this, take the derivative of $f(x)$ first, then substitute 3 in for $x$: $f'(x)=2x+3 \big\rvert_3=9$. Thus, the slope of the tangent line is 9.

  2. Again, simply find the value of the derivative function at $x=2$: $N(x)={(x^6-x^5)}^{\frac{1}{5}} \implies$ $ N'(x)=\frac{1}{5}(x^6-x^5)^{-\frac{4}{5}}\cdot(6x^5-5x^4) \implies$ $N'(x)\big\rvert_2=\frac{1}{5}\cdot\frac{1}{16}\cdot112=\frac{112}{80}=\frac{7}{5}$. This is the slope of the tangent line.

  3. We are evaluating $y'$ at $x=3$: $y'=2x \implies 2x\big\rvert_3=6$.

  4. In order to get the equation of the tangent line, you must first find its slope (the function's derivative evaluated at $x=\frac{\pi}{4}$):$y'=2\sin x \cos x \implies y'\big\rvert_\frac{\pi}{4}=2\left(\frac{1}{\sqrt2}\right)^2=1$. Next, substitute $\frac{\pi}{4}$ back into the original equation (not the derivative): $y\big\rvert_\frac{\pi}{4}=\left(\frac{1}{\sqrt2}\right)^2=\frac{1}{2}$. Now that we have a point $\left(\frac{\pi}{4},\frac{1}{2}\right)$ and a slope of 1, we can build the tangent line using point-slope form: $y-\frac{1}{2}=1\left(x-\frac{\pi}{4}\right)$. This is a perfectly acceptable equation of a tangent line.

  5. Since parallel lines have the same slope, we can deduce that the derivative at that particular $x$-value will be equal to 12. Expressing this algebraically, we find $y'=6x^2-18x-12=12 \implies$ $x^2-3x-4=0 \implies (x-4)(x+1)=0$. Since we are looking for a positive $x$-value, the answer will be $x=4$.

  6. It should be fairly obvious that they want us to find the equation of the tangent line. Luckily, we know how to do this. Yay! Take the derivative at $x=1$ to find the slope: $3x^2+2\big\rvert_1=5$. Next, plug $x=1$ back into the original function to find the point: $x^3+2x\big\rvert_1=3$. Point: (1,3), slope: 5. Plug into point-slope equation and simplify: $y-3=5(x-1) \implies y=5x+2$ Finally simply locate $A$ and $B$ and sum their squares: $5^2+2^2=29$.

  7. This is impossible because the domain of any $\log$ function is $\mathbb{N}$. DNE.

  8. Deriving this is simply a matter of implicit differentiation: $\frac{dy}{dx}(3x^2y-4x^3=1)\implies6xy+y’3x^2-12x^2=0$. Now all that’s left is to solve for $y’$: $y’=\frac{12x^2-6xy}{3x^2}$. To find the slope of the tangent line, all you have to do now is plug in $x=2$ into the equation: $y’\biggr\rvert_2=\frac{48-12(\frac{25}{12})}{12}=\frac{48-25}{12}=\frac{23}{12}$.

  9. By the definition of the chain rule, we know that $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. In order for the tangent to be vertical, the derivative must be undefined at that point; this means that there must be a zero in the denominator. Thus, all we have to do is find $\frac{dx}{dt}$ and set it equal to zero to find the value of $t$ which gives you this: $\frac{dx}{dt}=2t+1=0\implies t=-\frac{1}{2}$.

  10. Again, from the chain rule, we know that $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Evaluating this, we get $\frac{dy}{dx}=\frac{1+e^t}{2t+1}$ In order to understand which value of $t$ to plug in, we must plug in the given point $(0,1)$ to the parametric functions and set up a system of equations: $\begin{array}{ll} 0=&t^2+t\\ 1=&t+e^t \end{array}$ From this, we can deduce that $t=0$ when $x=0$ and $y=1$. Thus, we can substitute 0 into our derivative to get the slope of the tangent line: $\frac{1+e^t}{2t+1}\biggr\rvert_0=2$.

  11. First, rearrange the function to determine what $x$-value to use: $x^3+2x^2-x-2=0 \implies x=1$. We will be using this value for the rest of the problem, since it is the one we are concerned with (note: if you're confused as to why we couldn't just use the $y$ value, it's because the $y$ values change when taking the derivative. We are not concerned with what happens when $y=0$, but when $x=1$). From here, simply take the derivative and evaluate it at $x=1$ to find the slope of the tangent line: $f'(1)=3x^2+4x-1\big\rvert_1=6$. Next, find the $y$-value of the original function to complete the point: $f(1)=0$ (which was given). Now simply build your line through point-slope form: $y-0=6(x-1) \implies y=6x-6$.

  12. It would behoove you to always rearrange roots into their exponent forms: $\sqrt{x+1}=(x+)^{\frac{1}{2}}$. Now that that's done, we have to understand what the question is asking. "Find the linear approximation" means to find the value that the tangent line will have at that point (the tangent line is, after all, the instantaneous slope, and can be used to approximate values close to the point of reference). So basically, Find the equation of the tangent line at $x=0$: $f'(0)=\frac{1}{2}(x+1)^{-\frac{1}{2}} \big\rvert_0=\frac{1}{2}$. I hope you have noticed by now that our $x$ and $y$ values are the ordered pair (0,1). This is because we are taking the values near $x=0$, and since $f(0)=\sqrt1=1$, we are simply finding the value of the tangent line when $x=0.02$ ($\frac{1}{50}$). This brings us to: $y-1=\frac{1}{2}(x-0) \implies y= \frac{1}{2}x+1$. substituting in $x=\frac{1}{50}$, we get $y=\frac{1}{100}+\frac{100}{100} = \frac{101}{100} = 1.01$.

  13. We can obtain an estimate for a value of the function by plugging the value into the tangent line a tangent line function at a different point. Seeing that you need to construct a tangent line, you need $y$-values and the only perfect squares in the given domain are 9 and 16 themselves. Due to the nature of the square root graph (concave down), the largest over-approximation will occur at $x=9$. Thus, all we have to du is find the equation of the line tangent to the function $y=\sqrt{x}$ at $x=9, y=3$ and plug in 11 to get our estimate: $f'(9)=\frac{1}{2}9^{-\frac{1}{2}}=\frac{1}{2}\cdot\frac{1}{3}=\frac{1}{6}$. From this information, we know that the equation of the tangent line will be $(y-3)=\frac{1}{6}(x-9)\implies y=\frac{1}{6}x+\frac{9}{6}$. Now evaluating this line at $x=11$, we get our answer: $\frac{11}{6}+\frac{9}{6}=\frac{20}{6}=\frac{10}{3}$.

  14. Another trick question! This is impossible, since there is no such thing as the $\ln$ of a negative number. DNE.

Normal Lines

  1. Remember, the normal line is the line perpendicular to the tangent at the point of tangency, and since the two are perpendicular, their slopes will be the negative reciprocals of each other. The easiest way to do this is to find the slope of the tangent line, and then find its negative reciprocal. Slope of tangent = $y'\big\rvert_2=3x^2\big\rvert_2=12$ the negative reciprocal is $-\frac{1}{x}$ so the negative reciprocal of 12 would be $-\frac{1}{12}$. This is the slope of the normal line.

  2. (Note: this problem requires knowledge of logarithmic differentiation, don't worry if you haven't covered that yet). Same thing for this problem, find the negative reciprocal of the slope of the tangent: $f'(2)=2x2^x+x^22^x\ln2\big\rvert_2=16+16\ln2$= slope of tangent. Slope of normal line = $-\frac{1}{16+16\ln2}$.

  3. (This problem also requires you to know logarithmic differentiation). Find the slope of the tangent and normal lines and add them together. For tangent: $y'\big\rvert_0=\cos(\ln(x+1)+5x)\cdot(5+\frac{1}{(x+1)})\big\rvert_0=1\cdot6=6$. This means that the slope of the normal line is $-\frac{1}{6}$. Adding them together, you get $\frac{36}{6}-\frac{1}{6}=\frac{35}{6}$.

  4. In order to get the slope of the tangent line, we must first find $y'$ and evaluate it at the point $(0,1)$. This is just a matter of implicit differentiation and the product rule: $y'=1-\sin xy(y+xy')$(chain rule and product rule). Thus, $y'=1-y\sin xy-xy'\sin xy$ $\implies -1=-y'-xy'\sin xy-y\sin xy$ $=y'(-1-x\sin xy)-y\sin xy \implies$ $y'=\frac{-1+y\sin xy}{-1-x\sin xy}$. Now, plugging in our given point we get $\frac{-1+\sin 0}{-1-0}=\frac{-1}{-1}=1$. Since this is the slope of the tangent line, the slope of the normal line will be the negative reciprocal, which is $-1$.

  5. When dealing with prametric equations, always remember the chain rule: $\frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$. Since we need $\frac{dy}{dx}$ to solve for the normal line, we must first ealuate $\frac{dy}{dt}$ and $\frac{dx}{dt}$ separately: $\frac{dx}{dt}=8t-3$; to solve for $\frac{dy}{dt}$ we must use logarithmic differentiation: $\frac{y'}{y}=t\ln(6e+4)\implies y'=[t\ln(6e+4)][6e^t+4]$. From this, we get that $\frac{dy}{dx}=\frac{[t\ln(6e+4)][6e^t+4]}{8t-3}$. To find the slope of the tangent line, all thats left is plugging in $\ln4$ for $t$, which yields the answer of $\frac{28\ln^24}{8\ln4-3}$.

  6. This question is asking you to find the equations of two normal lines, equate them, solve the equation, and plug the solution back in to get your final answer. That seems like a lot of work! It's ok since we know what we're doing: you know that the slope of the normal line is the negative reciprocal of the slope of the tangent line. Working this out algebraically, we get that the slope of $L_1=2$ and the slope of $L_2=4$. This must mean that the slopes of the normal lines are $-\frac{1}{2}$ and $-\frac{1}{4}$ respectively. In order to find the whole equation of the normal lines we mus plug in $x=1$ and $x=2$ for $y=x^2$, yielding answers of 1 and 4. Now all we have to do is put it into slope-intercept form: $\begin{array}{lll} L_1:&(y-1)=-\frac{1}{2}(x-1)\implies&y=-\frac{1}{2}x+\frac{3}{2}\\ L_2:&(y-4)=-\frac{1}{4}(x-2)\implies&y=-\frac{1}{4}x+\frac{9}{2} \end{array}$. now that we have the lines, all that's left is to find their points of intersection, which is done by setting them equal to each other and solving for $x:-\frac{1}{2}x+\frac{3}{2}=-\frac{1}{4}x+\frac{9}{2}\implies-\frac{1}{4}x=3\implies$ $x=-12\implies y=\frac{15}{2}$.

  7. To solve for the normal lines, we must first solve for the tangent line, which can be obtained by solving for $y'$ via implicit differentiation. We also need to know what $y$-value to plug in. With some manipulation, thegiven equation can be made to look like this: $2y^2=y(7x^2+x^3)-8x\implies2y=x^3+7x^2-8x$. Now plugging in 1 for $x$, we get that $y=0$. Evaluating the derivative of the equation, we get the following: $4yy'=(7xy+yx^2-8)+x(7y7xy'+y'x^3+2xy)$ $=7xy+7x^2y'+y'x^3+2x^2y+7xy+yx^2-8$ $\implies4yy'-7x^2y'-y'x^3$ $=7xy+2x^2y+7xy+yx^2-8=$ $y'(4y-7x^2-x^3)\implies y'=\frac{7xy+2x^2y+7xy+yx^2-8}{4y-7x^2-x^3}$. Now all we have to do is plug in $x=1$ and $y=0$ to this equation to get the slope of the tangent line: $\frac{-8}{-8}=1$. Thus, the slope of the normal line is -1. The equation of the normal line can the be written as $(y-0)=-1(x-1)\implies y=-x+1$.

Tabular Derivatives

  1. $h(x)=f(x)g(x) \implies h'(x)=f'(x)g(x)+g'(x)f(x)$ (product rule) $f'(5)=4\cdot -2+5 \cdot 3 =7$.

  2. Quotient Rule: $h(x)=\frac{f(x)}{g(x)} \implies h'(4)=\frac{f'(4)g(4)-g'(4)f(4)}{(g(4))^2} =- \frac{9}{9} = -1$.

  3. You are given that $f'(x)g(x)+g'(x)f(x)=7$ (product rule expansions). Solving for $g(3)$, we get $5g(3)+4\cdot 4=7 \implies g(3)=-\frac{9}{5}$ now we simply multiply: $5\cdot -\frac{9}{5}=-9$.

  4. $(g(f(x))) \big\rvert_1=g'(f(x))\cdot f'(x)\big\rvert_1= g'(2)\cdot f'(1)=-12$.

  5. Given the definition of inverse fractions, we can set up the following equation: $h(f(x))=x$. Now differentiating, we get $h'(f(x))\cdot f'(x)=1 \implies h'(2) \cdot-2=1 \implies h'(2)=-\frac{1}{2}$.

  6. $h'(2)=g'(f(2))\cdot f'(2) = -3 \cdot 6 =-18$.

  7. We can again set up an equation knowing the definition of composite functions: $h(g(x))=x$. Differentiating, we get $h'(g(1))\cdot g'(1)=1$ Solving, $h'(2)\cdot -3=1 \implies h'(2)=-\frac{1}{3}$.

  8. Look at all this algebra: $h(x)=\frac{(g(x^2))^2}{(x+2)f(x)} \implies$ $h'(x)=\frac{(2(g(x^2))\cdot g'(x^2)\cdot 2x)((x+2)f(x))-((f(x)+f'(x)(x+2))(g(x^2))^2)}{((x+2)f(x))^2} \implies$ $h'(2)=\frac{((2)(-2)(-3)(4)(4)(\frac{1}{4}))-(((\frac{1}{4})+(2)(4))(4))}{((4)(\frac{1}{4}))^2}=\frac{48-33}{1}=15$.

  9. To solve this, first differentiate what you're given: $\frac{d}{dx} (f(g(x^2))=x) = (f'(g(x^2))\cdot (g'(x^2)) \cdot 2x =1)$ From this, we get that $g'(x^2)=\frac{1}{f'(g(x^2))(2x)}$. Now, since we are solving for $g'(x^2)$, and since the $x$ is the same throughout, we will use $x=2$ ($2^2=4$). Plugging this in to the original function yields $f(g(x^2))=2 \implies f(g(4))=2 \implies g(4)=-1$. Now that we have all the values, we simply plug them in to our derivative function: $g'(4)=\frac{1}{f'(-1)(2\cdot2)} = \frac{1}{3\cdot4}=\frac{1}{12}$.