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QuestionAssigament 3.1 Find the derivative without using a calculator.
1. y=(3x+5)3y=(3 x+5)^{3}
2. f(x)=3(7x+5)4f(x)=3(7 x+5)^{4}
3. y=2−3xy=\sqrt{2-3 x}
4. f(t)=1(1−t)2f(t)=\frac{1}{(1-t)^{2}}
5. y=(x2+1)2y=\sqrt{\left(x^{2}+1\right)^{2}}
6. g(x)=x(2x+3)3g(x)=x(2 x+3)^{3}
7. y=1x+1y=\frac{1}{\sqrt{x+1}}
8. f(x)=3x−2x+1f(x)=\frac{3 x-2}{x+1}
9. g(x)=sec⁡(4x)g(x)=\sec (4 x)
10. y=4tan⁡(2x)y=4 \tan (2 x)
11. f(θ)=12sin⁡2(3θ)f(\theta)=\frac{1}{2} \sin ^{2}(3 \theta)
12. y=4x3−2x2xy=\sqrt{\frac{4 x^{3}-2 x}{2 x}}
Find an equation of the line tangent to the graph of ff at the given point without using a calculator.
13. f(x)=2x2+2f(x)=\sqrt{2 x^{2}+2} at (−1,2)(-1,2)
15. f(x)=1(9x)3f(x)=\frac{1}{\sqrt{(9 x)^{3}}} at (17+527)\left(\frac{1}{7}+\frac{5}{27}\right)
14. f(x)=x+4xf(x)=\frac{x+4}{x} at (2,3)(2,3)
16. f(x)=1x2+cos⁡xf(x)=\frac{1}{x^{2}}+\sqrt{\cos x} at (2π,14π2+1)\left(2 \pi, \frac{1}{4 \pi^{2}}+1\right)
Find the indicated derivatives.
17. ddx(2sin⁡x−3)4\frac{d}{d x}(2 \sin x-3)^{4}
18. d2dt2(t2−1)32\frac{d^{2}}{d t^{2}}\left(t^{2}-1\right)^{\frac{3}{2}}
19. Find the point(s) at which a line tangent to the graph of f(x)=(2x−3)s f(x)=(2 x-3)^{\text {s }} is parallel to the graph of y=24x−7y=24 x-7. You may use a calculator.
20. If g(x)=(f(x))3,f(1)=2g(x)=(f(x))^{3}, f(1)=2, and f′(1)=4f^{\prime}(1)=4, find g′(1)g^{\prime}(1)
21. Given these values \begin{tabular}{|c|c|c|c|c|} \hlinexx & f(x)f(x) & g(x)g(x) & f′(x)f^{\prime}(x) & g′(x)g^{\prime}(x) \\ \hline 2 & 3 & 2 & -1 & 4 \\ \hline 3 & -2 & 12\frac{1}{2} & 6 & 5 \\ \hline \end{tabular} find the following derivatives. a. ddxg(f(x))\frac{d}{d x} g(f(x)) at x=2x=2 c. ddxg(x)\frac{d}{d x} \sqrt{g(x)} at x=2x=2 b. ddr(g(x)f(x))\frac{d}{d r}(g(x) f(x)) at x=2x=2 d. ddxg(x)f(x)\frac{d}{d x} \frac{g(x)}{f(x)} at x=2x=2

Studdy Solution

STEP 1

1. We assume familiarity with basic differentiation rules and techniques, such as the power rule, product rule, chain rule, and quotient rule.
2. We assume the ability to simplify algebraic expressions and apply trigonometric identities where necessary.
3. We assume the knowledge of how to find the equation of a tangent line using the point-slope form of a line.

STEP 2

1. Differentiate each given function using appropriate differentiation rules.
2. For problems involving the tangent line, find the derivative at the given point and use the point-slope form to find the equation of the tangent line.
3. For higher-order derivatives, apply the differentiation rules repeatedly.
4. For finding points where tangents are parallel, equate the derivative to the given slope and solve for the points.

STEP 3

Differentiate y=(3x+5)3 y = (3x + 5)^3 using the chain rule.
dydx=3⋅(3x+5)2⋅ddx(3x+5)=3⋅(3x+5)2⋅3=9(3x+5)2 \frac{dy}{dx} = 3 \cdot (3x + 5)^2 \cdot \frac{d}{dx}(3x + 5) = 3 \cdot (3x + 5)^2 \cdot 3 = 9(3x + 5)^2

STEP 4

Differentiate f(x)=3(7x+5)4 f(x) = 3(7x + 5)^4 using the chain rule.
dfdx=3⋅4(7x+5)3⋅ddx(7x+5)=3⋅4(7x+5)3⋅7=84(7x+5)3 \frac{df}{dx} = 3 \cdot 4(7x + 5)^3 \cdot \frac{d}{dx}(7x + 5) = 3 \cdot 4(7x + 5)^3 \cdot 7 = 84(7x + 5)^3

STEP 5

Differentiate y=2−3x y = \sqrt{2 - 3x} using the chain rule.
y=(2−3x)1/2 y = (2 - 3x)^{1/2} dydx=12(2−3x)−1/2⋅ddx(2−3x)=12(2−3x)−1/2⋅(−3)=−322−3x \frac{dy}{dx} = \frac{1}{2}(2 - 3x)^{-1/2} \cdot \frac{d}{dx}(2 - 3x) = \frac{1}{2}(2 - 3x)^{-1/2} \cdot (-3) = -\frac{3}{2\sqrt{2 - 3x}}

STEP 6

Differentiate f(t)=1(1−t)2 f(t) = \frac{1}{(1 - t)^2} using the chain rule.
f(t)=(1−t)−2 f(t) = (1 - t)^{-2} dfdt=−2(1−t)−3⋅ddx(1−t)=−2(1−t)−3⋅(−1)=2(1−t)3 \frac{df}{dt} = -2(1 - t)^{-3} \cdot \frac{d}{dx}(1 - t) = -2(1 - t)^{-3} \cdot (-1) = \frac{2}{(1 - t)^3}

STEP 7

Differentiate y=(x2+1)2 y = \sqrt{(x^2 + 1)^2} by simplifying first.
y=∣x2+1∣ y = |x^2 + 1| Since x2+1>0 x^2 + 1 > 0 for all x x , y=x2+1 y = x^2 + 1 dydx=ddx(x2+1)=2x \frac{dy}{dx} = \frac{d}{dx}(x^2 + 1) = 2x

STEP 8

Differentiate g(x)=x(2x+3)3 g(x) = x(2x + 3)^3 using the product rule.
dgdx=(2x+3)3+x⋅3(2x+3)2⋅ddx(2x+3) \frac{dg}{dx} = (2x + 3)^3 + x \cdot 3(2x + 3)^2 \cdot \frac{d}{dx}(2x + 3) dgdx=(2x+3)3+x⋅3(2x+3)2⋅2 \frac{dg}{dx} = (2x + 3)^3 + x \cdot 3(2x + 3)^2 \cdot 2 dgdx=(2x+3)3+6x(2x+3)2 \frac{dg}{dx} = (2x + 3)^3 + 6x(2x + 3)^2 dgdx=(2x+3)2[(2x+3)+6x] \frac{dg}{dx} = (2x + 3)^2 \left[(2x + 3) + 6x \right] dgdx=(2x+3)2[8x+3] \frac{dg}{dx} = (2x + 3)^2 \left[8x + 3 \right]

STEP 9

Differentiate y=1x+1 y = \frac{1}{\sqrt{x + 1}} using the chain rule.
y=(x+1)−1/2 y = (x + 1)^{-1/2} dydx=−12(x+1)−3/2 \frac{dy}{dx} = -\frac{1}{2}(x + 1)^{-3/2}

STEP 10

Differentiate f(x)=3x−2x+1 f(x) = \frac{3x - 2}{x + 1} using the quotient rule.
dfdx=(x+1)⋅3−(3x−2)⋅1(x+1)2 \frac{df}{dx} = \frac{(x + 1) \cdot 3 - (3x - 2) \cdot 1}{(x + 1)^2} dfdx=3(x+1)−(3x−2)(x+1)2 \frac{df}{dx} = \frac{3(x + 1) - (3x - 2)}{(x + 1)^2} dfdx=3x+3−3x+2(x+1)2 \frac{df}{dx} = \frac{3x + 3 - 3x + 2}{(x + 1)^2} dfdx=5(x+1)2 \frac{df}{dx} = \frac{5}{(x + 1)^2}

STEP 11

Differentiate g(x)=sec⁡(4x) g(x) = \sec(4x) using the chain rule.
dgdx=sec⁡(4x)⋅tan⁡(4x)⋅4 \frac{dg}{dx} = \sec(4x) \cdot \tan(4x) \cdot 4 dgdx=4sec⁡(4x)tan⁡(4x) \frac{dg}{dx} = 4 \sec(4x) \tan(4x)

STEP 12

Differentiate y=4tan⁡(2x) y = 4 \tan(2x) using the chain rule.
dydx=4⋅sec⁡2(2x)⋅ddx(2x) \frac{dy}{dx} = 4 \cdot \sec^2(2x) \cdot \frac{d}{dx}(2x) dydx=4⋅sec⁡2(2x)⋅2 \frac{dy}{dx} = 4 \cdot \sec^2(2x) \cdot 2 dydx=8sec⁡2(2x) \frac{dy}{dx} = 8 \sec^2(2x)

STEP 13

Differentiate f(θ)=12sin⁡2(3θ) f(\theta) = \frac{1}{2} \sin^2(3\theta) using the chain rule.
dfdθ=12⋅2sin⁡(3θ)⋅cos⁡(3θ)⋅ddθ(3θ) \frac{df}{d\theta} = \frac{1}{2} \cdot 2 \sin(3\theta) \cdot \cos(3\theta) \cdot \frac{d}{d\theta}(3\theta) dfdθ=sin⁡(3θ)⋅cos⁡(3θ)⋅3 \frac{df}{d\theta} = \sin(3\theta) \cdot \cos(3\theta) \cdot 3 dfdθ=3sin⁡(3θ)cos⁡(3θ) \frac{df}{d\theta} = 3 \sin(3\theta) \cos(3\theta) dfdθ=32sin⁡(6θ) \frac{df}{d\theta} = \frac{3}{2} \sin(6\theta)

STEP 14

Differentiate y=4x3−2x2x y = \sqrt{\frac{4x^3 - 2x}{2x}} by simplifying first.
y=2x2−1 y = \sqrt{2x^2 - 1} y=(2x2−1)1/2 y = (2x^2 - 1)^{1/2} dydx=12(2x2−1)−1/2⋅ddx(2x2−1) \frac{dy}{dx} = \frac{1}{2}(2x^2 - 1)^{-1/2} \cdot \frac{d}{dx}(2x^2 - 1) dydx=12(2x2−1)−1/2⋅4x \frac{dy}{dx} = \frac{1}{2}(2x^2 - 1)^{-1/2} \cdot 4x dydx=2x2x2−1 \frac{dy}{dx} = \frac{2x}{\sqrt{2x^2 - 1}}

STEP 15

Find the equation of the tangent line for f(x)=2x2+2 f(x) = \sqrt{2x^2 + 2} at (−1,2) (-1, 2) .
First, differentiate f(x)=(2x2+2)1/2 f(x) = (2x^2 + 2)^{1/2} : f′(x)=12(2x2+2)−1/2⋅4x=2x2x2+2 f'(x) = \frac{1}{2}(2x^2 + 2)^{-1/2} \cdot 4x = \frac{2x}{\sqrt{2x^2 + 2}}
Evaluate f′(−1) f'(-1) : f′(−1)=2(−1)2(−1)2+2=−24=−1 f'(-1) = \frac{2(-1)}{\sqrt{2(-1)^2 + 2}} = \frac{-2}{\sqrt{4}} = -1
Using the point-slope form y−y1=m(x−x1) y - y_1 = m(x - x_1) : y−2=−1(x+1) y - 2 = -1(x + 1) y=−x+1 y = -x + 1
The equation of the tangent line is y=−x+1 y = -x + 1 .

STEP 16

Find the equation of the tangent line for f(x)=x+4x f(x) = \frac{x + 4}{x} at (2,3) (2, 3) .
First, differentiate f(x)=x+4x f(x) = \frac{x + 4}{x} : f(x)=1+4x f(x) = 1 + \frac{4}{x} f′(x)=−4x2 f'(x) = -\frac{4}{x^2}
Evaluate f′(2) f'(2) : f′(2)=−422=−1 f'(2) = -\frac{4}{2^2} = -1
Using the point-slope form y−y1=m(x−x1) y - y_1 = m(x - x_1) : y−3=−1(x−2) y - 3 = -1(x - 2) y=−x+5 y = -x + 5
The equation of the tangent line is y=−x+5 y = -x + 5 .

STEP 17

Find the equation of the tangent line for f(x)=1(9x)3 f(x) = \frac{1}{\sqrt{(9x)^3}} at (17,527) \left( \frac{1}{7}, \frac{5}{27} \right) .
First, rewrite f(x) f(x) : f(x)=(9x)−3/2 f(x) = (9x)^{-3/2}
Differentiate f(x) f(x) : f′(x)=−32(9x)−5/2⋅9=−272(9x)−5/2 f'(x) = -\frac{3}{2} (9x)^{-5/2} \cdot 9 = -\frac{27}{2} (9x)^{-5/2}
Evaluate f′(17) f' \left( \frac{1}{7} \right) : f′(17)=−272(9⋅17)−5/2=−272(97)−5/2 f'\left( \frac{1}{7} \right) = -\frac{27}{2} \left( 9 \cdot \frac{1}{7} \right)^{-5/2} = -\frac{27}{2} \left( \frac{9}{7} \right)^{-5/2}
Simplify: f′(17)=−272(79)5/2=−272⋅(79)5/2 f'\left( \frac{1}{7} \right) = -\frac{27}{2} \left( \frac{7}{9} \right)^{5/2} = -\frac{27}{2} \cdot \left( \frac{7}{9} \right)^{5/2}
Using the point-slope form y−y1=m(x−x1) y - y_1 = m(x - x_1) with m=f′(17) m = f'\left( \frac{1}{7} \right) : y−527=−272(79)5/2(x−17) y - \frac{5}{27} = -\frac{27}{2} \left( \frac{7}{9} \right)^{5/2} \left( x - \frac{1}{7} \right)
The equation of the tangent line is: y−527=−272(79)5/2(x−17) y - \frac{5}{27} = -\frac{27}{2} \left( \frac{7}{9} \right)^{5/2} \left( x - \frac{1}{7} \right)

STEP 18

Find the equation of the tangent line for f(x)=1x2+cos⁡x f(x) = \frac{1}{x^2} + \sqrt{\cos x} at (2π,14π2+1) \left( 2\pi, \frac{1}{4\pi^2} + 1 \right) .
First, differentiate f(x) f(x) : f(x)=x−2+(cos⁡x)1/2 f(x) = x^{-2} + (\cos x)^{1/2} f′(x)=−2x−3−12(cos⁡x)−1/2⋅sin⁡x f'(x) = -2x^{-3} - \frac{1}{2} (\cos x)^{-1/2} \cdot \sin x
Evaluate f′(2π) f'(2\pi) : f′(2π)=−2(2π)−3−12(cos⁡2π)−1/2⋅sin⁡2π f'(2\pi) = -2(2\pi)^{-3} - \frac{1}{2} (\cos 2\pi)^{-1/2} \cdot \sin 2\pi f′(2π)=−2(2π)3−12(1)−1/2⋅0 f'(2\pi) = -\frac{2}{(2\pi)^3} - \frac{1}{2} (1)^{-1/2} \cdot 0 f′(2π)=−28π3=−14π3 f'(2\pi) = -\frac{2}{8\pi^3} = -\frac{1}{4\pi^3}
Using the point-slope form y−y1=m(x−x1) y - y_1 = m(x - x_1) : y−(14π2+1)=−14π3(x−2π) y - \left( \frac{1}{4\pi^2} + 1 \right) = -\frac{1}{4\pi^3} (x - 2\pi)
The equation of the tangent line is: y=−14π3(x−2π)+(14π2+1) y = -\frac{1}{4\pi^3} (x - 2\pi) + \left( \frac{1}{4\pi^2} + 1 \right)

STEP 19

Find the derivative ddx(2sin⁡x−3)4 \frac{d}{dx} (2 \sin x - 3)^4 using the chain rule.
ddx(2sin⁡x−3)4=4(2sin⁡x−3)3⋅ddx(2sin⁡x−3) \frac{d}{dx} (2 \sin x - 3)^4 = 4(2 \sin x - 3)^3 \cdot \frac{d}{dx}(2 \sin x - 3) =4(2sin⁡x−3)3⋅2cos⁡x = 4(2 \sin x - 3)^3 \cdot 2 \cos x =8(2sin⁡x−3)3cos⁡x = 8(2 \sin x - 3)^3 \cos x

STEP 20

Find the second derivative d2dt2(t2−1)3/2 \frac{d^2}{dt^2} \left( t^2 - 1 \right)^{3/2} .
First, find the first derivative: ddt(t2−1)3/2=32(t2−1)1/2⋅2t=3t(t2−1)1/2 \frac{d}{dt} \left( t^2 - 1 \right)^{3/2} = \frac{3}{2} \left( t^2 - 1 \right)^{1/2} \cdot 2t = 3t \left( t^2 - 1 \right)^{1/2}
Next, find the second derivative using the product rule: d2dt2=ddt(3t(t2−1)1/2) \frac{d^2}{dt^2} = \frac{d}{dt} \left( 3t \left( t^2 - 1 \right)^{1/2} \right) =3(t2−1)1/2+3t⋅ddt((t2−1)1/2) = 3 \left( t^2 - 1 \right)^{1/2} + 3t \cdot \frac{d}{dt} \left( \left( t^2 - 1 \right)^{1/2} \right) =3(t2−1)1/2+3t⋅12(t2−1)−1/2⋅2t = 3 \left( t^2 - 1 \right)^{1/2} + 3t \cdot \frac{1}{2} \left( t^2 - 1 \right)^{-1/2} \cdot 2t =3(t2−1)1/2+3t2(t2−1)−1/2 = 3 \left( t^2 - 1 \right)^{1/2} + 3t^2 \left( t^2 - 1 \right)^{-1/2} =3(t2−1)−1/2(t2−1+t2) = 3 \left( t^2 - 1 \right)^{-1/2} \left( t^2 - 1 + t^2 \right) =3(t2−1)−1/2(2t2−1) = 3 \left( t^2 - 1 \right)^{-1/2} \left( 2t^2 - 1 \right)

STEP 21

Find the point(s) at which a line tangent to the graph of f(x)=(2x−3)5 f(x) = (2x - 3)^5 is parallel to the graph of y=24x−7 y = 24x - 7 .
First, find the derivative of f(x) f(x) : f′(x)=5(2x−3)4⋅2=10(2x−3)4 f'(x) = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4
We need to find x x such that f′(x)=24 f'(x) = 24 : 10(2x−3)4=24 10(2x - 3)^4 = 24 (2x−3)4=2410=2.4 (2x - 3)^4 = \frac{24}{10} = 2.4 2x−3=±2.44 2x - 3 = \pm \sqrt[4]{2.4} 2x=3±2.44 2x = 3 \pm \sqrt[4]{2.4} x=3±2.442 x = \frac{3 \pm \sqrt[4]{2.4}}{2}
The points are: \[ x = \frac{3 + \sqrt[4

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