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3 years ago

1_1/2 + 1_3/. find the sum

Mathematics

1 answer:

vesna_86 [32]3 years ago

8 0

Answer:

-1/5

Step-by-step explanation:

Your question is a little badly written . But if take it properly , The solution will be like :

1-0/5+1-3=-1-0/5=-1/5

Good luck ^^

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Differentiate x^2 e^x logx

zimovet [89]

Product rule:

$\dfrac{\mathrm d}{\mathrm dx}(x^2e^x\log x)$

$=\dfrac{\mathrm d(x^2)}{\mathrm dx}e^x\log x+x^2\dfrac{\mathrm d(e^x)}{\mathrm dx}\log x+x^2e^x\dfrac{\mathrm d(\log x)}{\mathrm dx}$

Power rule:

$\dfrac{\mathrm d(x^2)}{\mathrm dx}=2x$

The exponential function is its own derivative:

$\dfrac{\mathrm d(e^x)}{\mathrm dx}=e^x$

Assuming the base of $\log x$ is $e$ , its derivative is

$\dfrac{\mathrm d(\log x)}{\mathrm dx}=\dfrac1x$

But if you mean a logarithm of arbitrary base $b$ , we have

$y=\log_bx\implies x=b^y=e^{y\ln b}\implies1=e^{y\ln b}\ln b\dfrac{\mathrm dy}{\mathrm dx}$

$\implies\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{e^{-y\ln b}}{\ln b}=\dfrac1{b^y\ln b}$

$\implies\dfrac{\mathrm d(\log_bx)}{\mathrm dx}=\dfrac1{x\ln b}$

So we end up with

$2xe^x\log x+x^2e^x\log x+\dfrac{x^2e^x}x$

$=xe^x(2\log x+x\log x+1)$

8 0

3 years ago

Evaluate g(x)=6x when x=-3,0, and 4 .

hram777 [196]

Answer:

-3 = -18

0 = 0

4 = 24

Step-by-step explanation:

-3... so 6(-3) equals -18 because a positive times a negative equals a negative

0... so 6 (0) equals 0 because anything multiplied by 0 equals 0

4... so 6 (4) equals 24

() means multiplied by so if it’s y(x) then that means y is multiplied by x :)

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Answer:

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marysya [2.9K]

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3 years ago

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JulijaS [17]

Answer:

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Step-by-step explanation:

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