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NCERT Exemplar · Q31

Q.Differentiate with respect to xx: (3x+5)(1+tan⁡x)(3x + 5)(1 + \tan x).

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Use the product rule: differentiate each factor in turn while holding the other constant, then add. The derivative is (3x+5)sec⁡2x+3(1+tan⁡x)\boxed{(3x + 5)\sec^2 x + 3(1 + \tan x)}.

When you have two functions multiplied together, you cannot simply differentiate each one separately and multiply the results. The product rule exists precisely because differentiation "sees" how both factors change and how they interact as xx changes.

Think of it this way: if u(x)u(x) grows while v(x)v(x) stays constant, the product u⋅vu \cdot v grows at rate u′⋅vu' \cdot v. If v(x)v(x) grows while u(x)u(x) stays constant, the product grows at rate u⋅v′u \cdot v'. Since both can change simultaneously, the total rate of change is the sum of these two contributions.

ddx[u(x)⋅v(x)]=u′(x)⋅v(x)+u(x)⋅v′(x)\frac{d}{dx}[u(x) \cdot v(x)] = u'(x) \cdot v(x) + u(x) \cdot v'(x)

Now let's apply this to (3x+5)(1+tan⁡x)(3x + 5)(1 + \tan x).

Step-by-step differentiation:

  1. Identify the two factors.

    Let u(x)=3x+5u(x) = 3x + 5 and v(x)=1+tan⁡xv(x) = 1 + \tan x.

  2. Differentiate the first factor.

    u′(x)=ddx(3x+5)=3u'(x) = \frac{d}{dx}(3x + 5) = 3.

  3. Differentiate the second factor.

    v′(x)=ddx(1+tan⁡x)=0+sec⁡2x=sec⁡2xv'(x) = \frac{d}{dx}(1 + \tan x) = 0 + \sec^2 x = \sec^2 x.

    Recall that the derivative of tan⁡x\tan x is sec⁡2x\sec^2 x, one of the standard trigonometric derivatives.

  4. Apply the product rule. …

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