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Continuity and Differentiability — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsContinuity and Differentiability2 Marks
Question
Differentiate $(x^2 - 5x + 8) (x^3 + 7x + 9)$ in three ways mentioned below:
  1. by using product rule
  2. by expanding the product to obtain a single polynomial
  3. by logarithmic differentiation.
    Do they all give the same answer?
✓

Answer

  1. Given: $(x^2 – 5x + 8) (x^3 + 7x + 9)$
    Let y $= (x^2 – 5x + 8) (x^3 + 7x + 9)$
    By applying product rule differentiate both sides with respect to $x$
    $\frac{d y}{d x}=\frac{d}{d x} (x^2 − 5x + 8)(x^3 + 7x + 9)$
    $\Rightarrow \frac{d y}{d x}=\left(x^{3}+7 x+9\right) \cdot \frac{d}{d x}\left(x^{2}-5 x+8\right)$ + $\left(x^{2}-5 x+8\right) \cdot \frac{d}{d x}\left(x^{3}+7 x+9\right)$
    $\Rightarrow \frac{d y}{d x}=\left(x^{3}+7 x+9\right) \cdot(2 x-5)+\left(x^{2}-5 x+8\right) \cdot\left(3 x^{2}+7\right)$ $\Rightarrow \frac{{dy}}{{dx}}=2{x}^{4} +14 {x}^{2}+18 \mathrm{x}-5{x}^{3} - 35 x  − 45 + 3 x ^4+ 7 x ^2− 15 x ^3− 35 x + 24 x ^2 + 56$
    $\Rightarrow \frac{d y}{d x}= 5x^4− 20x^3+ 45x^2− 52x + 11$
  2. Given:$ (x^2 – 5x + 8) (x^3 + 7x + 9)$
    Let $y = (x^2 – 5x + 8) (x^3 + 7x + 9)$
    $\Rightarrow y = (x^2 – 5x + 8) (x^3 + 7x + 9)$
    $\Rightarrow y = x^5 + 7x^3 + 9x^2 - 5x^4 – 35x^2 - 45x + 8x^3 + 56x + 72$
    $\Rightarrow y = x^5 - 5x^4 + 15x^3 - 26x^2 + 11x + 72$
    Now, differentiate both sides with respect to x, we get
    $\frac{d y}{d x}=\frac{d}{d x}\left(x^{5}\right)-\frac{d}{d x}\left(5 x^{4}\right)+\frac{d}{d x}\left(15 x^{3}\right)-\frac{d}{d x}\left(26 x^{2}\right)+\frac{d}{d x}(11 x)+\frac{d}{d x}(72)$
    $\frac{d y}{d x} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$
  3. Given: $(x^2 – 5x + 8) (x^3 + 7x + 9)$
    Let $y = (x^2 – 5x + 8) (x^3 + 7x + 9)$
    Taking log on both sides, we get
    $\log y = \log ((x^2 – 5x + 8) (x^3 + 7x + 9))$
    $\Rightarrow \log y = \log (x^2 – 5x + 8) + \log (x^3 + 7x + 9)$
    Now, differentiate both sides with respect to $x$
    $\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})=\frac{\mathrm{d}}{\mathrm{dx}} \log \left(\mathrm{x}^{2}-5 \mathrm{x}+8\right)+\frac{\mathrm{d}}{\mathrm{dx}} \log \left(\mathrm{x}^{3}+7 \mathrm{x}+9\right)$
    $\Rightarrow \frac{1}{\mathrm{y}} \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{y})=\left[\frac{1}{\left(\mathrm{x}^{2}-5 \mathrm{x}+8\right)} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{2}-5 \mathrm{x}+8\right)+\frac{1}{\left(\mathrm{x}^{3}+7 \mathrm{x}+9\right)} \cdot \frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{x}^{3}+7 \mathrm{x}+9\right)\right]$
    $\Rightarrow \frac{1}{y} \frac{d}{d x}(y)=\left[\frac{1}{\left(x^{2}-5 x+8\right)} \cdot(2 x-5)+\frac{1}{\left(x^{3}+7 x+9\right)} \cdot\left(3 x^{2}+7\right)\right]$
    $\Rightarrow \frac{d}{d x}(y)=y \cdot\left[\frac{(2 x-5)}{\left(x^{2}-5 x+8\right)}+\frac{\left(3 x^{2}+7\right)}{\left(x^{3}+7 x+9\right)}\right]$
    $\Rightarrow \frac{d}{d x}(y)=y \cdot\left[\frac{(2 x-5)\left(x^{3}+7 x+9\right)+\left(3 x^{2}+7\right)\left(x^{2}-5 x+8\right)}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$
    $\Rightarrow \frac{d}{d x}(y)$$=y \cdot\left[\frac{2 x^{4}+14 x^{2}+18 x-5 x^{3}-35 x-45+3 x^{4}-15 x^{3}+24 x^{2}+7 x^{2}-35 x+56}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$
    $\Rightarrow \frac{d}{d x}(y)=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$.$\left[\frac{5 x^{4}-20 x^{3}-45 x^{2}-52 x+11}{\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)}\right]$
    $\Rightarrow \frac{d y}{d x} = 5x^4 - 20x^3 + 45x^2 - 52x + 11$

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