Question

2log2x differentiate them

Answer 1

your answer for question

Answer 2

$\bf \orange{Question:}$

$\tt{How \: can \: I \: differentiate \: 2^{log}2x?}$

$\bf \orange{Solution:}$

*A2A.

Recall the formula:

$\tt{log_{e}xy=(\log_{e}x+\log_{e}y)}$

This,in turn,implies that:

$\tt{\log_{e}2x=\log_{e}2+\log_{e}x}$

Again,look at the formula:

$\tt{\log_{a}b=\log_{c}b\times \log_{a}c}$

This again gives the notion that:

$\tt{\log_{e}x=\log_{2}x\times \log_{e}2}$

Henceforth, $\tt{2^{\log_{e}2x}=2^{\log_{e}2+\log_{e}x}}$

$\tt{=2^{\log_{e}2}\cdot 2^{\log_{e}x}}$

$\tt{=2^{\log_{e}2}\cdot 2^{\log_{2}x\cdot \log_{e}2}}$

$\tt{=2^{\log_{e}2}\cdot (x^{\log_{e}2})}$

Since, $\log_{e}$2 is a constant,we can replace(or rather substitute) it with n.

$\tt{\implies \dfrac{\mathrm d(2^{\log_{e}2x})}{\mathrm dx}=\dfrac{\mathrm d(2^{n}x^{n})}{\mathrm dx}}$

$\tt{=2^{n}n(x^{n-1})}$

$\tt{=2^{\log_{e}2}\log_{e}2\cdot (x^{\log_{e}2-1})}$

$\boxed{\boxed{\dfrac{\mathrm d(2^{\log_{e}2x})}{\mathrm dx}=2^{\log_{e}2}\log_{e}2\cdot (x^{\log_{e}2-1})}}$

I hope it's helps you ☺️.

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