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Exercise 5.7 · Q5

Q.Find dydx\frac{dy}{dx} in the following: x3log⁡xx^3 \log x

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We differentiate x3log⁡xx^3 \log x using the product rule (since it is a product of x3x^3 and log⁡x\log x). The derivative is 3x2log⁡x+x23x^2 \log x + x^2.

The function given is y=x3log⁡xy = x^3 \log x. This is a product of two distinct functions: x3x^3 (a power function) and log⁡x\log x (the natural logarithm). Whenever you have a product of two functions, the natural tool is the product rule, not the chain rule. The chain rule would apply if we had a composition like log⁡(x3)\log(x^3) or (x3)2(x^3)^2, but here the functions are multiplied, not nested.

The product rule states: if y=u⋅vy = u \cdot v, then

dydx=udvdx+vdudx.\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}.

Let’s apply it step by step.

  1. Identify the two factors.

    Let u=x3u = x^3 and v=log⁡xv = \log x.

    (Here log⁡x\log x means the natural logarithm, base ee, as is standard in calculus.)

  2. Differentiate each factor separately.

    • Derivative of u=x3u = x^3:

dudx=3x2.\frac{du}{dx} = 3x^2.

  • Derivative of v=log⁡xv = \log x:

dvdx=1x.\frac{dv}{dx} = \frac{1}{x}.

  1. Apply the product rule.

dydx=udvdx+vdudx=x3⋅1x+log⁡x⋅3x2.\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} = x^3 \cdot \frac{1}{x} + \log x \cdot 3x^2.

  1. Simplify the first term. x3⋅1x=x2x^3 \cdot \frac{1}{x} = x^2. So we have:

dydx=x2+3x2log⁡x.\frac{dy}{dx} = x^2 + 3x^2 \log x.

  1. Factor if desired (optional). Both terms share a factor x2x^2, so we can write: …

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