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Using Product rule, find the derivatives of the following functions: (i) y = (3x2+2)(5x−3). (ii) y = ex sin x. (iii) y = log x.

Mathematics

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3 parts
i

45x2−18x+10\mathbf{45x^2 - 18x + 10}

ii

e2x(2sin⁡x+cos⁡x)\mathbf{e^{2x}(2\sin x + \cos x)}

iii

e2x(2log⁡x+1x)\mathbf{e^{2x}\left(2\log x + \frac{1}{x}\right)}

Step-by-step solution

12 steps

Here are the derivatives using the product rule for each function:

Part i· 4 steps

Given f(x)=(3x2+2)(5x−3)f(x) = (3x^2 + 2)(5x - 3)

  1. Identify uu and vv.

    Let u=3x2+2u = 3x^2 + 2 and v=5x−3v = 5x - 3.

  2. Find the derivatives of uu and vv.

    u′=ddx(3x2+2)=6xu' = \frac{d}{dx}(3x^2 + 2) = 6x
    v′=ddx(5x−3)=5v' = \frac{d}{dx}(5x - 3) = 5

  3. Apply the product rule f′(x)=u′v+uv′f'(x) = u'v + uv'.

    f′(x)=(6x)(5x−3)+(3x2+2)(5)f'(x) = (6x)(5x - 3) + (3x^2 + 2)(5)
  4. Expand and simplify the expression.

    f′(x)=30x2−18x+15x2+10f'(x) = 30x^2 - 18x + 15x^2 + 10 f′(x)=45x2−18x+10f'(x) = 45x^2 - 18x + 10

    The derivative is 45x2−18x+10\boxed{\mathbf{45x^2 - 18x + 10}}.

Part ii

e2x(2sin⁡x+cos⁡x)\mathbf{e^{2x}(2\sin x + \cos x)}

Given y=e2xsin⁡xy = e^{2x} \sin x

  1. Identify uu and vv.

    Let u=e2xu = e^{2x} and v=sin⁡xv = \sin x.

  2. Find the derivatives of uu and vv.

    u′=ddx(e2x)=2e2xu' = \frac{d}{dx}(e^{2x}) = 2e^{2x}
    v′=ddx(sin⁡x)=cos⁡xv' = \frac{d}{dx}(\sin x) = \cos x

  3. Apply the product rule dydx=u′v+uv′\frac{dy}{dx} = u'v + uv'.

    dydx=(2e2x)(sin⁡x)+(e2x)(cos⁡x)\frac{dy}{dx} = (2e^{2x})(\sin x) + (e^{2x})(\cos x)
  4. Factor out common terms.

    dydx=e2x(2sin⁡x+cos⁡x)\frac{dy}{dx} = e^{2x}(2\sin x + \cos x)

    The derivative is e2x(2sin⁡x+cos⁡x)\boxed{\mathbf{e^{2x}(2\sin x + \cos x)}}.

Part iii

e2x(2log⁡x+1x)\mathbf{e^{2x}\left(2\log x + \frac{1}{x}\right)}

Given y=e2xlog⁡xy = e^{2x} \log x (assuming log⁡x\log x refers to the natural logarithm, ln⁡x\ln x)

  1. Identify uu and vv.

    Let u=e2xu = e^{2x} and v=log⁡xv = \log x.

  2. Find the derivatives of uu and vv.

    u′=ddx(e2x)=2e2xu' = \frac{d}{dx}(e^{2x}) = 2e^{2x}
    v′=ddx(log⁡x)=1xv' = \frac{d}{dx}(\log x) = \frac{1}{x}

  3. Apply the product rule dydx=u′v+uv′\frac{dy}{dx} = u'v + uv'.

    dydx=(2e2x)(log⁡x)+(e2x)(1x)\frac{dy}{dx} = (2e^{2x})(\log x) + (e^{2x})\left(\frac{1}{x}\right)
  4. Factor out common terms.

    dydx=e2x(2log⁡x+1x)\frac{dy}{dx} = e^{2x}\left(2\log x + \frac{1}{x}\right)

    The derivative is e2x(2log⁡x+1x)\boxed{\mathbf{e^{2x}\left(2\log x + \frac{1}{x}\right)}}.

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