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aniked [119]
3 years ago
13

Please find the derivative of

Bx%5E2%7D" id="TexFormula1" title="\displaystyle \frac{e^{\frac{3}{x}}}{x^2}" alt="\displaystyle \frac{e^{\frac{3}{x}}}{x^2}" align="absmiddle" class="latex-formula">. Show all work necessary - thanks!
Mathematics
2 answers:
sesenic [268]3 years ago
7 0

Hello! :)

\large\boxed{\frac{-e^{\frac{3}{x}}  (3 + 2x )}{x^{4}}}

Find the derivative using the quotient rule:

\frac{f(x)}{g(x)} = \frac{g(x) * f'(x) - f(x) * g'(x)}{(g(x))^{2}}

In this instance:

f(x) = e^{\frac{3}{x} }\\\\g(x) = x^{2}

Use the following properties to find the derivative of f(x) and g(x):

e^{u} = u' * e^{u}\\\\x^{n} = nx^{n-1}

Use the quotient rule:

\frac{x^{2} * (e^{\frac{3}{x}} * (-3x^{-2})) - e^{\frac{3}{x}} * 2x  }{(x^{2} )^{2}}

Simplify the numerator:

\frac{(e^{\frac{3}{x}} * (-3)) - e^{\frac{3}{x}} * 2x  }{(x^{2} )^{2}}

Factor out e^{\frac{3}{x}}

\frac{e^{\frac{3}{x}}  (-3 - 2x )}{x^{4}}

Factor out -1 from the numerator:

\frac{-e^{\frac{3}{x}}  (3 + 2x )}{x^{4}}

And we're done! Thanks for posting the question to my 1000th answer!

jekas [21]3 years ago
6 0

Answer:

\displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}}}{x^4} - \frac{2e^{\frac{3}{x}}}{x^3}

General Formulas and Concepts:

<u>Pre-Algebra</u>

  • Splitting Fractions

<u>Algebra I</u>

  • Terms/Coefficients
  • Factoring
  • Exponential Rule [Multiplying]: \displaystyle b^m \cdot b^n = b^{m + n}

<u>Calculus</u>

Derivatives

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

Derivative Rule:    \displaystyle \frac{d}{dx} [e^u]=e^u \cdot u'

Quotient Rule:      \displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}

Step-by-step explanation:

<u>Step 1: Define</u>

<u />\displaystyle \frac{e^{\frac{3}{x}}}{x^2}\\f(x) = e^{\frac{3}{x}}\\g(x) = x^2<u />

<u />

<u>Step 2: Differentiate</u>

  1. Quotient Rule:                                                                                                \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{\frac{d}{dx}[e^{\frac{3}{x}}] \cdot x^2 - \frac{d}{dx}[x^2] \cdot e^{\frac{3}{x}}}{(x^2)^2}
  2. Derivative Rule:                                                                                              \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{e^{\frac{3}{x}} \cdot \frac{-3}{x^2} \cdot x^2 - \frac{d}{dx}[x^2] \cdot e^{\frac{3}{x}}}{(x^2)^2}
  3. [Simplify] Multiply:                                                                                           \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}} - \frac{d}{dx}[x^2] \cdot e^{\frac{3}{x}}}{(x^2)^2}
  4. Basic Power Rule:                                                                                          \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}} - 2x^{2-1} \cdot e^{\frac{3}{x}}}{(x^2)^2}
  5. [Simplify] Subtract Exponents:                                                                       \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}} - 2x \cdot e^{\frac{3}{x}}}{(x^2)^2}
  6. [Simplify] Multiply:                                                                                           \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}} - 2xe^{\frac{3}{x}}}{(x^2)^2}
  7. [Simplify] Exponent Rule:                                                                               \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}} - 2xe^{\frac{3}{x}}}{x^{2 + 2}}
  8. [Simplify] Add Exponents:                                                                              \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}} - 2xe^{\frac{3}{x}}}{x^4}
  9. [Simplify] Fraction Split:                                                                                  \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}}}{x^4} - \frac{2xe^{\frac{3}{x}}}{x^4}
  10. [Simplify - 2nd Fraction] Cancel Like Terms:                                                \displaystyle \frac{d}{dx}[\frac{e^{\frac{3}{x}}}{x^2}] = \frac{-3e^{\frac{3}{x}}}{x^4} - \frac{2e^{\frac{3}{x}}}{x^3}

And we have our final answer!

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Use the method from the Example to approximate the solution to the equations below to two decimal places.
shutvik [7]

Answer:

The approximate solution of the equation is 4,29.

Step-by-step explanation:

The solve of this fomula is:

5ˣ = 1000

You can use log in each side of the formula::

log₁₀ 5ˣ = log₁₀ 1000

By logarithm law (logₐ (mⁿ) = n logₐ m. And as log₁₀(1000) = 3:

x log₁₀ 5 = 3

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I hope it helps!

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3 years ago
There are 7 cows and 5 chicken in the farmer's filed
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Answer:

7:12

Step-by-step explanation:

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8 0
3 years ago
15+10•8 div by 4 must equal 35
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Suppose a random variable X has the following probability density function:f(x)=1/x, 1≤x≤C, or f(x)=0 otherwise a) what must the
Sergeeva-Olga [200]

Answer:

a)  C=e^1=e

b) P(X

c) E(X) =\int_{1}^e x \frac{1}{x} dx =x \Big|_1^e \ =e-1

E(X^2) =\int_{1}^e x^2 \frac{1}{x} dx =\frac{x^2}{2} \Big|_1^e \ =\frac{1}{2}(e^2 -1)

Var(X)=E(X^2)-[E(X)]^2= \frac{1}{2}(e^2 -1) -(e-1)^2 = 0.242

Step-by-step explanation:

a) what must the value of C be so that f(x) is a probability density function?

In order to be a probability function we need this condition:

\int_{1}^C \frac{1}{x} dx =1

And solving the left part of the integral we have:

ln(x) \Big|_1^C \ =1

ln(C)-ln(1)=1, so then C=e^1=e

b) find P(X<2)

We can find this probability on this way using the density function:

P(X

c) find E(X) and Var(X)

We can find the expected value on this way:

E(X) =\int_{1}^e x \frac{1}{x} dx =x \Big|_1^e \ =e-1

In order to find the Var(X) we need to find the second moment given by:

E(X^2) =\int_{1}^e x^2 \frac{1}{x} dx =\frac{x^2}{2} \Big|_1^e \ =\frac{1}{2}(e^2 -1)

And now we can use the following definition:

Var(X)=E(X^2)-[E(X)]^2= \frac{1}{2}(e^2 -1) -(e-1)^2 = 0.242

5 0
4 years ago
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White raven [17]

Answer:

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Step-by-step explanation:

Pic below...

hope this helped!

brainliest please!

7 0
3 years ago
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