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To find the derivative (dy)/(dx) of the given function, we will differentiate each term separately.

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To find the derivative dydx\frac{dy}{dx} of the given function, we will differentiate each term separately.

The given function is:

y=5x3−2ln⁡(5x)+3x3+5xy = \frac{5}{x^3} - 2 \ln(5x) + \frac{3}{\sqrt[3]{x}} + \sqrt{5x}

First, rewrite the terms using exponent notation:

y=5x−3−2ln⁡(5x)+3x−1/3+(5x)1/2y = 5x^{-3} - 2 \ln(5x) + 3x^{-1/3} + (5x)^{1/2}

Now, we differentiate each term:

  1. Differentiate the first term, 5x−35x^{-3}.

    Using the power rule ddx(cxn)=cnxn−1\frac{d}{dx}(cx^n) = cnx^{n-1}:

    ddx(5x−3)=5(−3)x−3−1=−15x−4=−15x4\frac{d}{dx}(5x^{-3}) = 5(-3)x^{-3-1} = -15x^{-4} = -\frac{15}{x^4}

  2. Differentiate the second term, −2ln⁡(5x)-2 \ln(5x).

    Using the chain rule for ln⁡(u)\ln(u), where u=5xu=5x and dudx=5\frac{du}{dx}=5:

    ddx(−2ln⁡(5x))=−2⋅15x⋅5=−2⋅55x=−2⋅1x=−2x\frac{d}{dx}(-2 \ln(5x)) = -2 \cdot \frac{1}{5x} \cdot 5 = -2 \cdot \frac{5}{5x} = -2 \cdot \frac{1}{x} = -\frac{2}{x}

  3. Differentiate the third term, 3x−1/33x^{-1/3}.

    Using the power rule:

    ddx(3x−1/3)=3(−13)x−1/3−1=−1x−4/3=−1x4/3\frac{d}{dx}(3x^{-1/3}) = 3\left(-\frac{1}{3}\right)x^{-1/3-1} = -1x^{-4/3} = -\frac{1}{x^{4/3}}

  4. Differentiate the fourth term, (5x)1/2(5x)^{1/2}.

    Using the chain rule for unu^n, where u=5xu=5x and dudx=5\frac{du}{dx}=5:

    ddx((5x)1/2)=12(5x)1/2−1⋅5=12(5x)−1/2⋅5=525x\frac{d}{dx}((5x)^{1/2}) = \frac{1}{2}(5x)^{1/2-1} \cdot 5 = \frac{1}{2}(5x)^{-1/2} \cdot 5 = \frac{5}{2\sqrt{5x}}

    We can simplify 525x\frac{5}{2\sqrt{5x}}:

    525x=525x=5525x=52x\frac{5}{2\sqrt{5x}} = \frac{5}{2\sqrt{5}\sqrt{x}} = \frac{\sqrt{5}\sqrt{5}}{2\sqrt{5}\sqrt{x}} = \frac{\sqrt{5}}{2\sqrt{x}}

  5. Combine all the derivatives.

    dydx=−15x4−2x−1x4/3+52x\frac{dy}{dx} = -\frac{15}{x^4} - \frac{2}{x} - \frac{1}{x^{4/3}} + \frac{\sqrt{5}}{2\sqrt{x}}

    The final answer is −15x4−2x−1x4/3+52x\boxed{-\frac{15}{x^4} - \frac{2}{x} - \frac{1}{x^{4/3}} + \frac{\sqrt{5}}{2\sqrt{x}}}.

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