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goldenfox [79]
2 years ago
9

Find first derivative f(x)= x² - 4 /x³ + 9 ​

Mathematics
2 answers:
Damm [24]2 years ago
7 0

Answer:

f'(x)=2x+\dfrac{12}{x^4}

Step-by-step explanation:

f(x)=x^2-\dfrac{4}{x^3}+9

Apply exponent rule \dfrac{1}{a^b}=a^{-b}:

\implies f(x)=x^2-4x^{-3}+9

Differentiate using the power rule \frac{d}{dx}(x^a)=a \cdot x^{a-1} :

\implies f'(x)=2 \cdot x^{2-1}-(-3)4x^{-3-1}+0

\implies f'(x)=2x+12x^{-4}

\implies f'(x)=2x+\dfrac{12}{x^4}

vitfil [10]2 years ago
5 0

Answer:

f'(x) = 2x + \frac{12}{x^{4} }

Step-by-step explanation:

differentiate using the power rule

\frac{d}{dx} ( ax^{n} ) = nax^{n-1}

Given

f(x) = x² - \frac{4}{x^3} + 9 = x² - 4x^{-3} + 9 , then

f'(x) = 2x + 12x^{-4} = 2x + \frac{12}{x^{4} }

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Lim n→∞[(n + n² + n³ + .... nⁿ)/(1ⁿ + 2ⁿ + 3ⁿ +....nⁿ)]​
Schach [20]

Step-by-step explanation:

\large\underline{\sf{Solution-}}

Given expression is

\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} }

To, evaluate this limit, let we simplify numerator and denominator individually.

So, Consider Numerator

\rm :\longmapsto\:n +  {n}^{2} +  {n}^{3}  +  -  -  -  +  {n}^{n}

Clearly, if forms a Geometric progression with first term n and common ratio n respectively.

So, using Sum of n terms of GP, we get

\rm \:  =  \: \dfrac{n( {n}^{n}  - 1)}{n - 1}

\rm \:  =  \: \dfrac{ {n}^{n}  - 1}{1 -  \dfrac{1}{n} }

Now, Consider Denominator, we have

\rm :\longmapsto\: {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  -  +  {n}^{n}

can be rewritten as

\rm :\longmapsto\: {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  -  +  {(n - 1)}^{n} +   {n}^{n}

\rm \:  =  \:  {n}^{n}\bigg[1 +\bigg[{\dfrac{n - 1}{n}\bigg]}^{n} + \bigg[{\dfrac{n - 2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]

\rm \:  =  \:  {n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]

Now, Consider

\rm :\longmapsto\:\displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} }

So, on substituting the values evaluated above, we get

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{\dfrac{ {n}^{n}  - 1}{1 -  \dfrac{1}{n} }}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{ {n}^{n}  - 1}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{ {n}^{n}\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{{n}^{n}\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{\bigg[1 - \dfrac{1}{ {n}^{n} } \bigg]}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

\rm \:  =  \: \displaystyle\lim_{n \to \infty}  \frac{1}{\bigg[1 +\bigg[1 - {\dfrac{1}{n}\bigg]}^{n} + \bigg[1 - {\dfrac{2}{n}\bigg]}^{n} +  -  -  -  + \bigg[{\dfrac{1}{n}\bigg]}^{n} \bigg]}

Now, we know that,

\red{\rm :\longmapsto\:\boxed{\tt{ \displaystyle\lim_{x \to \infty} \bigg[1 + \dfrac{k}{x} \bigg]^{x}  =  {e}^{k}}}}

So, using this, we get

\rm \:  =  \: \dfrac{1}{1 +  {e}^{ - 1}  + {e}^{ - 2} +  -  -  -  -  \infty }

Now, in denominator, its an infinite GP series with common ratio 1/e ( < 1 ) and first term 1, so using sum to infinite GP series, we have

\rm \:  =  \: \dfrac{1}{\dfrac{1}{1 - \dfrac{1}{e} } }

\rm \:  =  \: \dfrac{1}{\dfrac{1}{ \dfrac{e - 1}{e} } }

\rm \:  =  \: \dfrac{1}{\dfrac{e}{e - 1} }

\rm \:  =  \: \dfrac{e - 1}{e}

\rm \:  =  \: 1 - \dfrac{1}{e}

Hence,

\boxed{\tt{ \displaystyle\lim_{n \to \infty}  \frac{n +  {n}^{2}  +  {n}^{3}  +  -  -  +  {n}^{n} }{ {1}^{n} +  {2}^{n} +  {3}^{n}  +  -  -  +  {n}^{n} } =  \frac{e - 1}{e} = 1 -  \frac{1}{e}}}

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