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Define a substitution for the inner function.

Computer Science

ScanSolve AI Answer

Answer

A

Step-by-step solution

5 steps

To find dydx\frac{dy}{dx} for y=log⁡e(1x)y = \log_e \left(\frac{1}{x}\right), we will use the chain rule.

  1. Define a substitution for the inner function.

    Let u=1xu = \frac{1}{x}.
    Then y=log⁡euy = \log_e u.

  2. Differentiate yy with respect to uu.

    dydu=ddu(log⁡eu)=1u\frac{dy}{du} = \frac{d}{du}(\log_e u) = \frac{1}{u}

  3. Differentiate uu with respect to xx.

    We can write u=x−1u = x^{-1}.

    dudx=ddx(x−1)=−1⋅x−1−1=−x−2=−1x2\frac{du}{dx} = \frac{d}{dx}(x^{-1}) = -1 \cdot x^{-1-1} = -x^{-2} = -\frac{1}{x^2}

  4. Apply the chain rule formula dydx=dydu×dudx\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}.

    dydx=(1u)×(−1x2)\frac{dy}{dx} = \left(\frac{1}{u}\right) \times \left(-\frac{1}{x^2}\right)

  5. Substitute u=1xu = \frac{1}{x} back into the expression for dydx\frac{dy}{dx}.

    dydx=(11x)×(−1x2)\frac{dy}{dx} = \left(\frac{1}{\frac{1}{x}}\right) \times \left(-\frac{1}{x^2}\right) dydx=(x)×(−1x2)\frac{dy}{dx} = (x) \times \left(-\frac{1}{x^2}\right) dydx=−xx2\frac{dy}{dx} = -\frac{x}{x^2} dydx=−1x\frac{dy}{dx} = -\frac{1}{x}

    Comparing this result with the given options, the correct option is A.

    The final answer is A. −1x\boxed{\text{A. }-\frac{1}{x}}.

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