How do I find partial derivatives with fractional exponents in the Chain Rule?

[harpazo]

Given k = ge^(x/y), find ∂k/∂x and ∂k/∂y. The fractional exponent throws me into a loop. Can someone show me step by step how to tackle this problem?

 

[Prove It]

Given k = ge^(x/y), find ∂k/∂x and ∂k/∂y. The fractional exponent throws me into a loop. Can someone show me step by step how to tackle this problem?

If you can do $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left( x\right) \end{align*}$ and $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}y}\,\left( \frac{1}{y} \right) \end{align*}$ there is no reason you can't do this problem.

 

[harpazo]

If you can do $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left( x\right) \end{align*}$ and $\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}y}\,\left( \frac{1}{y} \right) \end{align*}$ there is no reason you can't do this problem.

d/dx [x] = 1

d/dy [1/y]

We know that 1/y can be expressed as y^(-1).

So, d/dy [y^(-1)] = -y^(-2), which of course is -1/y^2.

 

[harpazo]

Given k = g*e^(x/y), find ∂k/∂x.

∂k/∂x = g * e^(x/y) * (1/y) + e^(x/y)

∂k/∂x = [g*e^(x/y)]/(y) + e^(x/y)

Note: When taking the derivative of the exponent (x/y), I first expressed (x/y) as x*(1/y) and then took the partial derivative of x holding 1/y as constant. The product rule is also at work here. Is this correct?

 

[harpazo]

Given k = g*e^(x/y), find ∂k/∂y.

∂k/∂y = g * e^(x/y) * (-x/y^2) + e^(x/y)

I applied the product rule.

 

[Prove It]

Given k = g*e^(x/y), find ∂k/∂x.

∂k/∂x = g * e^(x/y) * (1/y) + e^(x/y)

∂k/∂x = [g*e^(x/y)]/(y) + e^(x/y)

Note: When taking the derivative of the exponent (x/y), I first expressed (x/y) as x*(1/y) and then took the partial derivative of x holding 1/y as constant. The product rule is also at work here. Is this correct?

If g is a constant then you don't need the product rule. If g is a function of x, then yes, the product rule will be needed. In either case, you have applied it incorrectly, as you should not have that extra "+ e^(x/y)"

- - - Updated - - -

Given k = g*e^(x/y), find ∂k/∂y.

∂k/∂y = g * e^(x/y) * (-x/y^2) + e^(x/y)

I applied the product rule.

Again, if g is a constant, you should not have used the product rule. If g is a function of y, then you would have needed the product rule, but you would have done so incorrectly, as again you should not have "+ e^(x/y)".

 

[harpazo]

Given k = g*e^(x/y), find ∂k/∂x.

∂k/∂x = g * e^(x/y) * (1/y)

∂k/∂x = [g*e^(x/y)]/(y)

Am I ok now?

Note: Here, y is held as constant.

- - - Updated - - -

Given k = g*e^(x/y), find ∂k/∂y.

∂k/∂y = g * e^(x/y) * (-x/y^2)

Am I ok now?

Note: I could also simplify further by placing y^2 in the denominator. Here, x is held as constant.

 

[Prove It]

Given k = g*e^(x/y), find ∂k/∂x.

∂k/∂x = g * e^(x/y) * (1/y)

∂k/∂x = [g*e^(x/y)]/(y)

Am I ok now?

Note: Here, y is held as constant.

- - - Updated - - -

Given k = g*e^(x/y), find ∂k/∂y.

∂k/∂y = g * e^(x/y) * (-x/y^2)

Am I ok now?

Note: I could also simplify further by placing y^2 in the denominator. Here, x is held as constant.

Yes they're both fine now :)

 

[harpazo]

Yes they're both fine now :)

I find partial derivatives easier than calculus 1 derivatives.

 

[topsquark]

I find partial derivatives easier than calculus 1 derivatives.

The honeymoon isn't over yet. The partial derivatives act in the same way as the ordinary derivatives. It's just that your course hasn't given you the harder problems while you are learning new methods.

-Dan