I need help answering all parts of question #5. Please make sure to really explain and show steps so I understand how to do it. (5) Consider the following table of values for differentiable functions f andg.x f(x) f'(x) g(x) g(x)120.912-87.12/32e-8-3-7331.21-1TT +eFor each function given by y below, use the table and derivative rules to compute the desiredderivative.(a) If y(x) = f(x)+ g(x) – 5, finddyat x = 3.dx(b) If y(x) = f(x)g(x), find the slope of the line tangent to the graph of y at x = 2.(c) If y(x) = (f (x))/(g(x)), find the instantaneous rate of change of y when x = T.(d) If y(x) = (g(x))², do you think y is increasing, decreasing or neither at x = e? Explainyour answer.

(5)(a)If y(x)=f(x)+g(x), find dydx at x=3y(x)=f(x)+g(x)=dydx=f'(x)+g'(x) [By sum rule of…

Answered: (5) Consider the following table of…

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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I need help answering all parts of question #5. Please make sure to really explain and show steps so I understand how to do it.

(5) Consider the following table of values for differentiable functions f and
g.
x f(x) f'(x) g(x) g(x)
1
2
0.9
1
2
-8
7.1
2/3
2
e
-8
-3
-7
3
3
1.2
1
-1
TT +e
For each function given by y below, use the table and derivative rules to compute the desired
derivative.
(a) If y(x) = f(x)+ g(x) – 5, find
dy
at x = 3.
dx
(b) If y(x) = f(x)g(x), find the slope of the line tangent to the graph of y at x = 2.
(c) If y(x) = (f (x))/(g(x)), find the instantaneous rate of change of y when x = T.
(d) If y(x) = (g(x))², do you think y is increasing, decreasing or neither at x = e? Explain
your answer.

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Step 1

$(5) (a) I f y (x) = f (x) + g (x), f i n d \frac{d y}{d x} a t x = 3 y (x) = f (x) + g (x) = \frac{d y}{d x} = f' (x) + g' (x) [B y s u m r u l e o f d e r i v a t i v e s] = {(\frac{d y}{d x})}_{x = 3} = {[f' (x) + g' (x)]}_{x = 3} = {(\frac{d y}{d x})}_{x = 3} = f' (3) + g' (3) {(\frac{d y}{d x})}_{x = 3} = 2 + π [u \sin g t a b l e] b) y = f (x) . g (x) s l o p e t o y (x) a t x = 2 i s g i v e n b y \frac{d y}{d x} a t x = 2 y = f (x) . g (x) \Rightarrow \frac{d y}{d x} = f (x) . g' (x) + f' (x) . g (x) [P r o d u c t r u l e] \Rightarrow {(\frac{d y}{d x})}_{x = 2} = f (2) . g' (2) + f' (2) . g (2) = - 8 * \frac{2}{3} + 7.1 * \frac{2}{3} = \frac{- 16}{3} + \frac{14.2}{3} = \frac{- 1.8}{3} \Rightarrow {(\frac{d y}{d x})}_{x = 2} = - 0.6 s o, s l o p e o f \tan g e n t t o y a t x = 2 = - 0.6$

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