Differentiate
y=e^[p](p+p \sqrt{p})
2 answers:
Answer:
Step-by-step explanation:
We are given the following information in the question:
![y = e^p(p+p\sqrt{p})\\\\y = e^p(p + p^{\frac{3}{2}})](https://tex.z-dn.net/?f=y%20%3D%20e%5Ep%28p%2Bp%5Csqrt%7Bp%7D%29%5C%5C%5C%5Cy%20%3D%20e%5Ep%28p%20%2B%20p%5E%7B%5Cfrac%7B3%7D%7B2%7D%7D%29)
We will use the product rule to differentiate the above expression.
The product rule says that:
![\displaystyle\frac{d(uv)}{dx} = u.\frac{dv}{dx} + v \frac{du}{dx}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bd%28uv%29%7D%7Bdx%7D%20%3D%20u.%5Cfrac%7Bdv%7D%7Bdx%7D%20%2B%20v%20%5Cfrac%7Bdu%7D%7Bdx%7D)
The differentiation is done in the following ways:
![\displaystyle\frac{dy}{dx} = \frac{d(e^p)}{dp}(p+p\sqrt{p}) + e^p\frac{d(p+p\sqrt{p})}{dp}\\\\= e^p(p+p\sqrt{p}) + e^p(1+\frac{3}{2}\sqrt{p})\\\\= e^p(p+p\sqrt{p}+1+\frac{3}{2}\sqrt{p})](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Cfrac%7Bdy%7D%7Bdx%7D%20%3D%20%5Cfrac%7Bd%28e%5Ep%29%7D%7Bdp%7D%28p%2Bp%5Csqrt%7Bp%7D%29%20%2B%20e%5Ep%5Cfrac%7Bd%28p%2Bp%5Csqrt%7Bp%7D%29%7D%7Bdp%7D%5C%5C%5C%5C%3D%20e%5Ep%28p%2Bp%5Csqrt%7Bp%7D%29%20%2B%20%20e%5Ep%281%2B%5Cfrac%7B3%7D%7B2%7D%5Csqrt%7Bp%7D%29%5C%5C%5C%5C%3D%20e%5Ep%28p%2Bp%5Csqrt%7Bp%7D%2B1%2B%5Cfrac%7B3%7D%7B2%7D%5Csqrt%7Bp%7D%29)
The above is the required differentiation of the given expression.
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Step-by-step explanation:
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Step-by-step explanation:
Answer:
106 1/7
Step-by-step explanation:
The line ...
ī = 106.142857143
tells you the mean of the dataset is 106 1/7.
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