Math  /  Calculus

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=23xy=\sqrt{2-3 x}
4. f(t)=1(1t)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)=3x2x+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)=12sin2(3θ)f(\theta)=\frac{1}{2} \sin ^{2}(3 \theta)
12. y=4x32x2xy=\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+cosxf(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(2sinx3)4\frac{d}{d x}(2 \sin x-3)^{4}
18. d2dt2(t21)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)=(2x3)f(x)=(2 x-3)^{\text {s }} is parallel to the graph of y=24x7y=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
Find the point(s) at which a line tangent to the graph of f(x)=(2x3)5 f(x) = (2x - 3)^5 is parallel to the graph of y=24x7 y = 24x - 7 .
First, find the derivative of f(x) f(x) : f(x)=5(2x3)42=10(2x3)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(2x3)4=24 10(2x - 3)^4 = 24 (2x3)4=2410=2.4 (2x - 3)^4 = \frac{24}{10} = 2.4 2x3=±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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