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Dmitrij [34]
3 years ago
9

Find the derivative of y = sin(ln(5x2 − 2x))

Engineering
1 answer:
pickupchik [31]3 years ago
3 0

Answer:

y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)

Explanation:

Let y = \sin[\ln(5\cdot x^{2}-2\cdot x)] and we proceed to find the derivative by the following steps:

1) y = \sin[\ln(5\cdot x^{2}-2\cdot x)] Given

2) y = \sin [\ln[x\cdot (5\cdot x - 2)]] Distributive property

3) y = \sin[\ln x + \ln (5\cdot x - 2 )] \ln (a\cdot b) = \ln a + \ln b

4) y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)  \frac{d}{dx} (\sin x) = \cos x/\frac{d}{dx}(\ln x) = \frac{1}{x}/\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}/Rule of chain/Result

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Solution :

Given :

Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa

The phase diagram is provided below.

a). The phase is a standard mixture.

b). At pressure, p = 200 kPa, T = $T_{saturated}$

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c). Specific volume

  $v_{f}= 0.001061, \ \ v_g=0.88578 \ m^3/kg$

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d). Specific energy (u_x)

    $u_f=504.5 \ kJ/kg, \ \ u_{fg}=2024.6 \ kJ/kg$

   $u_x=504.5 + 0.7(2024.6)$

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e). Specific enthalpy $(h_x)$

   At $h_f = 504.71, \ \ h_{fg} = 2201.6$

   h_x=504.71+(0.7\times 2201.6)

        $= 2045.83 \ kJ/kg$

f). Enthalpy at m = 0.5 kg

  $H=mh_x$

       $= 0.5 \times 2045.83$

       = 1022.91 kJ

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