- #1
Altami
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Homework Statement
How would you differentiate, (v^3-2v*squareroot v)?
Homework Equations
The Attempt at a Solution
would it look like...3v^2 - v^-1/2 ?
Altami said:would it look like...3v^2 - v^-1/2 ?
micromass said:The 3v^2 part is correct! But how did you obtain v^-1/2??
micromass said:Well, [tex]\sqrt{x}=x^{1/2}[/tex], so you just need to apply the power rule!
But the problem is that you don't only have a square root, you have [tex]v\sqrt{v}[/tex]. To derive this, you have to apply the chain rule. (or if you know your algebra, you could notice that [tex]v^{3/2}=v\sqrt{v}[/tex] and immediately apply the power rule)...
micromass said:Can't you just cancel v from the numerator and denominator??
Altami said:y=(v^3 - 2v*square rootv)/ v
and I have to differentiate it, I know I have to use the quotient rule and I have tired but the dang square root is cause me problems.
Mark44 said:No, you don't have to use the quotient rule if you simplify this first.
[tex]y = \frac{v^3 - 2v\sqrt{v}}{v} = v^2 - 2\sqrt{v} = v^2 - 2v^{1/2}[/tex]
Mark44 said:Well, if the instructions are that you have to use the quotient rule, then I haven't been any help at all. If so, I would write the numerator as v3 - 2v3/2.
Differentiation is a mathematical process used to find the rate of change of a function with respect to its input variables. It involves calculating the derivative of a function, which gives the slope of the function at a specific point.
The main purpose of differentiation is to analyze the behavior of a function and understand how it changes over its input variables. It is also used to find maximum and minimum values of a function, and to solve optimization problems in various fields such as physics, economics, and engineering.
The general formula for differentiating a function f(x) is given by f'(x) = lim (h->0) [f(x+h) - f(x)] / h. This can also be written as f'(x) = df(x)/dx.
Yes, differentiation can be done on functions with multiple variables, such as f(x,y,z). In this case, partial derivatives are used to find the rate of change of the function with respect to each of its variables.
To differentiate a function with a radical term, such as v^3 - 2v*squareroot(v), we can use the power rule for differentiation, which states that the derivative of x^n is nx^(n-1). In this case, the derivative of v^3 would be 3v^2, and the derivative of 2v*squareroot(v) would be 2(v^(1/2))(1) + (2v)(1/2)(v^(-1/2)) = (2v^(1/2)) + (v^(-1/2)). Therefore, the derivative of the given function would be 3v^2 - (2v^(1/2)) - (v^(-1/2)).