some true and false about the chain rule ( I don't understand the notation or the question)

Question

part b) (fg)'(x) is never equal to f'(x)g'(x)

 

part d) if the derivative of f(g(x)) is equal to the derivative of f(x) for all of x, then g(x)=x for all x

 

Part e) suppose that f'' and g'' exist and that f and g are concave up for all x, then f(g(x)) is concave up for all x.

 

So I have to determine is these are true or false but I don't understand what they are saying. Please help me.

Answer 1

part b) The general formula for (fg)'(x) is (fg)'(x)=f'(x)g(x)+f(x)g'(x)

The question is asking whether or not it is possible to have an exception for this rule where (fg)'(x)=f'(x)g'(x). To satisfy this, let's equate the two:

 

f'(x)g(x)+f(x)g'(x) = f'(x)g'(x) (I will use f and g for simplicity for the next few equations)

 

rearranging, we get: fg' = f'(g' - g)

more rearranging: 

 

f/f' = (g' - g) / g' 

f/f' = 1 - g/g' and therefore, f/f' + g/g' = 1. Let's assume f(x) = g(x). We have: 2f/f' = 1 or, f' = 2f. This is possible if f(x) = e^(2x). 

 

Therefore, if f(x) = g(x) = e^(2x), (fg)'(x) = f'(x)g'(x) and the statement is FALSE. 

 

part d) The general formula for the derivative of f(g(x)) is  f(g(x))' = f'(g(x)).g'(x)

If we equate this to the case stated in the problem we get: f'(g(x)).g'(x) = f'(x). Therefore, 2 conditions must be met: 1. g(x) = x and g'(x) = 1; therefore, g(x) = x is the solution to the problem. (this is of course, assuming both f(x) and g(x) are non-constant. for example, assume f(x) = 5 and g(x) = 10). Then f'(x) = g'(x) = 0, also satisfying f(g(x))' = f'(x) = 0. Excluding the latter case, the statement is TRUE.

 

 

part e) The question is giving a condition: f'' and g'' are always concave up. Does this mean f(g(x)) is also always concave up as a result of both functions satisfying this condition? To answer this we know that a function is concave up if its second derivative is positive. We will find the second derivative of f(g(x)). I will use f and g for simplicity for the next few equations.

 

first derivative: f(g)' = f'(g)g'

second derivative: [f(g)']' = [f'(g).g']' = [f'(g)]'.g' + f'(g).[g']'  (we know (fg)' = f'g + fg')

 

Rearranging the above, we see: [f(g)']' = [f''(g).g'].g' + f'(g).g'' (note that in the first bracket we have to take the derivative of f'(g) which is f''(g).g'

 

Cleaning up: [f(g)']' = f''(g).(g'^2) + f'(g).g'' 

 

For f(g) to be concave, this equation must always be positive. We evaluate this equation: In the first term, f" and (g'^2) are positive. f'' because it's concave according to the question, and g'^2 because it's an even power function. On the second term, g'' is positive because g(x) is a concave up function, but we have no knowledge of f'(g). f'(x) could be either negative or positive. 

 

Therefore, the equation is the sum of an "always positive" term with a term which "may or may not  be positive". Therefore, we cannot guarantee that the second derivative of f(g(x)) is always positive and hence, the statement is FALSE. 

 

An example for this is f(x) = x^2 - x + 1 and g(x) = x^2

f''(x) = g''(x) = 2. However, let's evaluate f(g(x)) = (x^2)^2 - (x^2) + 1 = x^4 - x^2 + 1.

 

[f(g(x))]' = 4x^3 - 2x (first derivative)

[f(g(x))]'' = 12x^2 -2 (second derivative to evaluate)

 

h''(x) is negative for -0.41 < x < 0.41 and the graph is concave down for this range. 

Answer 2

Each of the three can be shown false by finding a counterexample.

 

b) False. Take g(x) = x. Then (fg)'(x) = f'(x) = f'(x)*1 = f'(x)g'(x)

 

The notation suggests that (b) is the product rule rather than the chain rule. Sometimes we use (fºg) and (f•g) to distinguish between function composition and function multiplication. If it is the latter, as the notation suggests, then taking either function equal to 0 shows (b) is false again, or taking both functions constant.

 

d) False. Take f constant. Then for any g(x), d/dx f(g(x)) ( = (fg)'(x) ) = f'(g(x))g'(x) = 0

 

e) False. Take f(x) = x² and g(x) = x²-1. Then f''(x) = g''(x) = 2 > 0, so g and g are concave up everywhere. But f(g(x)) = (x²-1)² is not concave up everywhere:

 

 

d/dx f(g(x)) = f'(g(x))g'(x)

 

d²/dx²  f(g(x)) = d/dx f'(g(x))g'(x)

= f''(g(x))g'(x)g'(x) + f'(g(x))g''(x)

= 2*4x² + 2*(x²-1)*2

= 12x² - 4

< 0  on (-1/√3,1/√3)  ==> concave down on this interval.

 

Answer 3

Part b

 

(fg)'(x) is the product rule.

 

(fg)'(x) = f'(x)g(x) + f(x)g'(x)

 

So the fact the (fg)'(x) is never equal to f'(x)g'(x) is true.

 

 

Part d

 

The derivative of f(g(x)) is the derivative of f(x) using the chain rule, as g(x) as the inner function and f(x) as the outer function. 

 

For example,

 

f(x) = e^x         g(x) = (x + 2)

 

f(g(x)) = f(x + 2) = e^{(x + 2)}

 

 

If g(x)=x, then you would have

 

derivative of f(g(x)) = derivative of f(x)        for all values of x

 

True.

 

 

Part e

 

Take the example from the last part and find the second derivative of f(g(x)).

 

f(g(x)) = e^{(x + 2)}

 

 

Using second derivative test, is the derivative positive for all values of x?  If yes, then true.  If not, then false.  This is the part you have to work on yourself to find out.