Question 201027
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1)  If *[tex \Large u = f(x)] then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d\left(b^u\right)}{dx} = b^u\ln(b)\frac{du}{dx}]


So for *[tex \Large 7^2^x], *[tex \Large u = 2^x], so *[tex \Large \frac{du}{dx} = 2^x\ln(2)] then:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{d\left(7^2^x\right)}{dx} = \left(7^2^x\ln(7)\right)\left(2^x\ln(2)\right)]


2) You have the product of two functions, so use the Product Rule:

If *[tex \Large f(x) = g(x) \cdot h(x)] then *[tex \Large f'(x) = g(x)h'(x) + h(x)g'(x)]


Here, *[tex \Large g(x) = x^5], so *[tex \Large g'(x) = 5x^4] (Power Rule)


and *[tex \Large h(x) = 5^x], so *[tex \Large h'(x) = 5^x\ln(5)] (see Prob 1)


Putting it all together:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ y = f(x) = x^5\cdot 5^x]


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ f'(x) = x^5\cdot5^x\ln(5) + 5^x\cdot5x^4]



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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