harpazo
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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?
Harpazo said: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 said: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 said: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 said: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.
Harpazo said: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.
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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 said:Yes they're both fine now :)
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.Harpazo said:I find partial derivatives easier than calculus 1 derivatives.