Questions from General Calculus

Q: If f (x) and g (x) are differentiable

If f (x) and g (x) are differentiable functions, find g (x) if you know that f (x) = 1>x and d/dx f (g (x)) = (2x + 5)/(x2 + 5x – 4).

Q: If f (x) and g (x) are differentiable

If f (x) and g (x) are differentiable functions, such that f (1) = 2, f ‘(1) = 3, f ‘(5) = 4, g(1) = 5, g ’(1) = 6, g ‘(2) = 7, and g ‘(5) = 8, find d/dx f ( g (x)) |x=1.

Q: Consider the functions of Exercise 59. Find d/dx g

Consider the functions of Exercise 59. Find d/dx g (f (x)) |x=1. Exercise 59: If f (x) and g (x) are differentiable functions, such that f (1) = 2, f ‘(1) = 3, f ‘(5) = 4, g(1) = 5, g ’(1) = 6, g...

Q: After a computer software company went public, the price of one

After a computer software company went public, the price of one share of its stock fluctuated according to the graph in Fig. 1(a). The total worth of the company depended on the value of one share and...

Q: Differentiate the function. y = 1 / (√x +

Differentiate the function. y = 1 / (√x + 1)

Q: Refer to Exercise 61. Use the chain rule to find dW

Refer to Exercise 61. Use the chain rule to find dW/dt |t=1.5 and dW/dt |t=3.5 . Give an interpretation for these values. Exercise 61: After a computer software company went public, the price of one...

Q: Refer to Exercise 61. (a) Find dx/

Refer to Exercise 61. (a) Find dx/dt |t=2.5 and dx/dt |t=4. Give an interpretation for these values. (b) Use the chain rule to find dW/dt |t=2.5 and dW/dt |t=4. Give an interpretation for these values...

Q: Refer to Exercise 61. (a) What was the

Refer to Exercise 61. (a) What was the maximum value of the company during the first 6 months since it went public, and when was that maximum value attained? (b) Assuming that the value of one share w...

Q: In an expression of the form f (g (x)),

In an expression of the form f (g (x)), f (x) is called the outer function and g (x) is called the inner function. Give a written description of the chain rule using the words inner and outer.

Q: Show that d/dx (3x) |x=0

Show that d/dx (3x) |x=0 ≈ 1.1 by calculating the slope (3h – 1)/h of the secant line passing through the points (0, 1) and (h, 3h). Take h = .1, .01, and .001.