Stumped on a question

Daniel S

I'm stuck on a question (http://www.wolframalpha.com/input/?i=derivative+of+%28sinx%29^+%28x%2B1%29) and I was hoping someone could give me a pointer. I understand what's being done the whole way down apart from where the following came from:

( du^v)/( du) = u^(-1+v) v

and

( du^v)/( dv) = u^v log(u)

Could someone explain that bit to me?

Michael Collins

Those two formula follow from the standard rules of calculus.

Say we want to find the derivative of the following, with respect to 'u'

[latex] \displaystyle y = u^{v} [/latex]

[latex] \displaystyle \frac{\hbox{d}y}{\hbox{d}u} = v\cdot u^{v-1}[/latex]

This is just a rule from calculus (which can of course be dervied from first principles). Let's take a specific example, and it might make more sense

[latex] \displaystyle y = x^{5} [/latex]

[latex] \displaystyle \frac{\hbox{d}y}{\hbox{d}x} = 5\cdot x^{4}.[/latex]

For the second one, we want the derivative with respect to the power this time, i.e. with respect to 'v'

[latex] \displaystyle y = u^{v} [/latex]

To hand a variable power, we can re-write this as

[latex] \displaystyle \ln(y) = \ln(u^{v}) = v\ln(u) [/latex]

Now taking the derivative with respect to v, using implicit differentiation gives

[latex] \displaystyle \frac{1}{y}\frac{\hbox{d}y}{\hbox{d}v} = \ln(u)[/latex]

so, rearranging and subbing in y from above

[latex] \displaystyle \frac{\hbox{d}y}{\hbox{d}v} = u^{v}\ln(u).[/latex]

Hopefully this makes sense.

Daniel S

Michael Collins wrote: »

Those two formula follow from the standard rules of calculus.

Say we want to find the derivative of the following, with respect to 'u'

[latex] \displaystyle y = u^{v} [/latex]

[latex] \displaystyle \frac{\hbox{d}y}{\hbox{d}u} = v\cdot u^{v-1}[/latex]

This is just a rule from calculus (which can of course be dervied from first principles). Let's take a specific example, and it might make more sense

[latex] \displaystyle y = x^{5} [/latex]

[latex] \displaystyle \frac{\hbox{d}y}{\hbox{d}x} = 5\cdot x^{4}.[/latex]

For the second one, we want the derivative with respect to the power this time, i.e. with respect to 'v'

[latex] \displaystyle y = u^{v} [/latex]

To hand a variable power, we can re-write this as

[latex] \displaystyle \ln(y) = \ln(u^{v}) = v\ln(u) [/latex]

Now taking the derivative with respect to v, using implicit differentiation gives

[latex] \displaystyle \frac{1}{y}\frac{\hbox{d}y}{\hbox{d}v} = \ln(u)[/latex]

so, rearranging and subbing in y from above

[latex] \displaystyle \frac{\hbox{d}y}{\hbox{d}v} = u^{v}\ln(u).[/latex]

Hopefully this makes sense.

That's perfect! Thank you!