Is the Derivative of 5*sqrt[x] Correct Without Using the Chain Rule?

[A_Munk3y]

Homework Statement

5*sqrt[x]

The Attempt at a Solution

=>5*(x)^(1/2)
=>2.5x^(-1/2)

is this right?

Or do you use chain rule here?
like =>5*(x)^(1/2)
=>5(1/2)(x)^(-1/2)*1)
=>2.5x^(-1/2)*5

 

[Char. Limit]

The first one is right.

 

[A_Munk3y]

Thanks

 

[rygza]

I think combination of product rule and chain rule.
5 * d/dx x^(1/2) + x^(1/2) * d/dx 5

which is just 5 * d/dx x^(1/2)
(use the chain rule on x^(1/2))

 

[Deneb Cyg]

It seems redundant to use the product and chain rules together. For an equation like this one it is much simpler to just use the general power rule for derivatives:

\frac{d}{dx}x^r=rx^r-1

In general the chain and product rules are only used when there are distinct functions f(x) and g(x). Doing what rygza is suggesting (though it gives you the correct answer) assumes f(x)=5 and g(x)=x^1/2 for the product rule portion. But f'(x)=0. Then for the chain rule portion f(x)=5x^1/2 and g(x)=x. But g'(x)=1.

So in summary you just do a bunch of extra steps before ending up with d/dx 5x^1/2 which requires the power rule to solve (=2.5x^-1/2)

 

[rygza]

It seems redundant to use the product and chain rules together. For an equation like this one it is much simpler to just use the general power rule for derivatives:

\frac{d}{dx}x^r=rx^r-1

In general the chain and product rules are only used when there are distinct functions f(x) and g(x). Doing what rygza is suggesting (though it gives you the correct answer) assumes f(x)=5 and g(x)=x^1/2 for the product rule portion. But f'(x)=0. Then for the chain rule portion f(x)=5x^1/2 and g(x)=x. But g'(x)=1.

So in summary you just do a bunch of extra steps before ending up with d/dx 5x^1/2 which requires the power rule to solve (=2.5x^-1/2)

lol totally forgot about the power rule . Yes, this would be the best way to go

 

[cyby]

The chain rule would have been applied to "x" in \sqrt(x), so the first one is right.