Differentiation

Peyton Manning

Guys, my differentiation is rusty. What is the differentiation of

a) x/7

b) 1/4 √x

Archimedes

Archimedes wrote: »

a) x/7

Well, [latex]\displaystyle\frac{x}{7}[/latex] is the same thing as [latex]\displaystyle\frac{1}{7}\mbox{x}[/latex]. So, differentiate [latex]\displaystyle\mbox{x}[/latex] and, at the very end, multiply it by [latex]\displaystyle\frac{1}{7}[/latex]. So, what's the derivative of x with respect to x?

b) 1/4 √x

For this one, change the [latex]\displaystyle\sqrt{x}[/latex] into its indice form, i.e. [latex]x^{\frac{1}{2}}[/latex]. Now, differentiate [latex]x^{\frac{1}{2}}[/latex] and, again at the very end, multiply it by [latex]\displaystyle\frac{1}{4}[/latex].

When there's a constant (a number) in front of the variable (in both of these cases, x), you can, basically, ignore it until the end and then multiply it by the derivative of the x.

Thomas_S_Hunterson

if [latex]f(x) = x^n[/latex] then [latex]f'(x) = n x^{n-1}[/latex].
and
[latex]x=x^1[/latex]
[latex]\sqrt{x} = x^{1/2}[/latex]
/edit: beaten to it

gerry87

y= x/7
dy/dx = 1/7

for the 2nd one is sqrt(x) above or below the line?

y = (1/4) * x^.5
dy/dx = .5*(1/4) * x^(-.5)
dy/dx = 1/(8*sqrt(x))

You can do sums like this in wolfram alpha http://www.wolframalpha.com/input/?i=differentiate+y%3Dx%2F7

Peyton Manning

I send many thanks and best wishes in all your respective directions