Question

Which one of the following statements is false?

the derivative with respect to x of the quotient of the quantity g of x and f(x) equals the quotient f of x times g prime of x minus g of x times f prime of x and the square of f of x

If f and g are differentiable, then the derivative with respect to x of the product of f of x and g of x equals the product of f prime of x and g prime of x.

If f and g are differentiable, then the derivative with respect to x of the square root of f of x equals the quotient of f of x and 2 times the square root of f prime of x.

None of these statements are false.

Answer 1

the first one is the quotient rule and it is correct

the 2nd one is the wrong form of the product rule, it should be dy/dx f(x)g(x)=f'(x)g(x)+g'(x)f(x)


3rd is correct

answe ris 2nd one

Answer 2

Answer: Hello there! lets go one by one

the derivative with respect to x of the quotient of the quantity g of x and f(x) equals the quotient f of x times g prime of x minus g of x times f prime of x and the square of f of x .

this says that: $\frac{d(g(x)/f(x))}{dx} = \frac{f(x)*g'(x) - g(x)f'(x)}{f(x)^{2} }$

This is true! this is teh derivative of a quotient.

If f and g are differentiable, then the derivative with respect to x of the product of f of x and g of x equals the product of f prime of x and g prime of x.

this says that: $\frac{d(f(x)*g(x)}{dx} = f'(x)*g'(x)$

this is false, the derivative of the product is:

$\frac{d(f(x)*g(x)}{dx} = f'(x)*g(x) + f(x)*g'(x)$

derivative with respect to x of the square root of f of x equals the quotient of f of x and 2 times the square root of f prime of x.

this says:

$\frac{d(\sqrt{f(x)} }{dx}  = f(x)/2\sqrt{f'(x)}$

lets do the derivative us first! :D

$\frac{d(\sqrt{f(x)} }{dx} = \frac{1}{2}\frac{1}{\sqrt{f(x)} } *f'(x)$

wich is different, then this statment is false.