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Differentiate the following: (i) y = (x + cos x)(x - tan x) (ii) (x + sec x)(x - tan x) (iii) y = x7 log x (iv) y = x³ tan x

Mathematics

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Step-by-step solution

16 steps

Here are the differentiations for each part:

Part i· 4 steps

y=(x+cos⁡x)(x−tan⁡x)y = (x + \cos x)(x - \tan x)

  1. Identify uu and vv.

    Let u=x+cos⁡xu = x + \cos x and v=x−tan⁡xv = x - \tan x.

  2. Find the derivatives of uu and vv.

    u′=ddx(x+cos⁡x)=1−sin⁡xu' = \frac{d}{dx}(x + \cos x) = 1 - \sin x v′=ddx(x−tan⁡x)=1−sec⁡2xv' = \frac{d}{dx}(x - \tan x) = 1 - \sec^2 x

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

    dydx=(1−sin⁡x)(x−tan⁡x)+(x+cos⁡x)(1−sec⁡2x)\frac{dy}{dx} = (1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x)

  4. Expand and simplify.

    dydx=(x−tan⁡x−xsin⁡x+sin⁡xtan⁡x)+(x−xsec⁡2x+cos⁡x−cos⁡xsec⁡2x)\frac{dy}{dx} = (x - \tan x - x \sin x + \sin x \tan x) + (x - x \sec^2 x + \cos x - \cos x \sec^2 x)

    Since cos⁡xsec⁡2x=cos⁡x⋅1cos⁡2x=1cos⁡x=sec⁡x\cos x \sec^2 x = \cos x \cdot \frac{1}{\cos^2 x} = \frac{1}{\cos x} = \sec x,

    dydx=x−tan⁡x−xsin⁡x+sin⁡xtan⁡x+x−xsec⁡2x+cos⁡x−sec⁡x\frac{dy}{dx} = x - \tan x - x \sin x + \sin x \tan x + x - x \sec^2 x + \cos x - \sec x dydx=2x−tan⁡x−xsin⁡x+sin⁡xtan⁡x−xsec⁡2x+cos⁡x−sec⁡x\boxed{\frac{dy}{dx} = 2x - \tan x - x \sin x + \sin x \tan x - x \sec^2 x + \cos x - \sec x}

Part ii· 4 steps

y=(x+sec⁡x)(x−tan⁡x)y = (x + \sec x)(x - \tan x)

  1. Identify uu and vv.

    Let u=x+sec⁡xu = x + \sec x and v=x−tan⁡xv = x - \tan x.

  2. Find the derivatives of uu and vv.

    u′=ddx(x+sec⁡x)=1+sec⁡xtan⁡xu' = \frac{d}{dx}(x + \sec x) = 1 + \sec x \tan x v′=ddx(x−tan⁡x)=1−sec⁡2xv' = \frac{d}{dx}(x - \tan x) = 1 - \sec^2 x

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

    dydx=(1+sec⁡xtan⁡x)(x−tan⁡x)+(x+sec⁡x)(1−sec⁡2x)\frac{dy}{dx} = (1 + \sec x \tan x)(x - \tan x) + (x + \sec x)(1 - \sec^2 x)

  4. Expand and simplify.

    dydx=(x−tan⁡x+xsec⁡xtan⁡x−sec⁡xtan⁡2x)+(x−xsec⁡2x+sec⁡x−sec⁡3x)\frac{dy}{dx} = (x - \tan x + x \sec x \tan x - \sec x \tan^2 x) + (x - x \sec^2 x + \sec x - \sec^3 x) dydx=2x−tan⁡x+xsec⁡xtan⁡x−sec⁡xtan⁡2x−xsec⁡2x+sec⁡x−sec⁡3x\boxed{\frac{dy}{dx} = 2x - \tan x + x \sec x \tan x - \sec x \tan^2 x - x \sec^2 x + \sec x - \sec^3 x}

Part iii· 4 steps

y=x7log⁡xy = x^7 \log x

  1. Identify uu and vv.

    Let u=x7u = x^7 and v=log⁡xv = \log x (assuming natural logarithm, ln⁡x\ln x).

  2. Find the derivatives of uu and vv.

    u′=ddx(x7)=7x6u' = \frac{d}{dx}(x^7) = 7x^6 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=(7x6)(log⁡x)+(x7)(1x)\frac{dy}{dx} = (7x^6)(\log x) + (x^7)\left(\frac{1}{x}\right)

  4. Simplify the expression.

    dydx=7x6log⁡x+x7−1\frac{dy}{dx} = 7x^6 \log x + x^{7-1} dydx=7x6log⁡x+x6\frac{dy}{dx} = 7x^6 \log x + x^6 dydx=x6(7log⁡x+1)\boxed{\frac{dy}{dx} = x^6(7 \log x + 1)}

Part iv· 4 steps

y=x3tan⁡xy = x^3 \tan x

  1. Identify uu and vv.

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

  2. Find the derivatives of uu and vv.

    u′=ddx(x3)=3x2u' = \frac{d}{dx}(x^3) = 3x^2 v′=ddx(tan⁡x)=sec⁡2xv' = \frac{d}{dx}(\tan x) = \sec^2 x

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

    dydx=(3x2)(tan⁡x)+(x3)(sec⁡2x)\frac{dy}{dx} = (3x^2)(\tan x) + (x^3)(\sec^2 x)

  4. Simplify the expression.

    dydx=3x2tan⁡x+x3sec⁡2x\boxed{\frac{dy}{dx} = 3x^2 \tan x + x^3 \sec^2 x}
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