Question

(i) Differentiate $(x^2-5x+8)(x^3+7x+9)$ by product rule.

(ii) Differentiate $(x^2-5x+8)(x^3+7x+9)$ by expanding the product to obtain a single polynomial.

(iii) Differentiate $(x^2-5x+8)(x^3+7x+9)$ by logarithmic differentiation.

Answer 1

(i) Given : $y=(x^2-5x+8)(x^3+7x+9)$

Differentiating both sides w.r.t. $x$ , we get,

$\frac{dy}{dx}=\frac{d(x^2-5x+8)(x^3+7x+9)}{dx}$

$\Rightarrow \frac{dy}{dx}=(x^3+7x+9)\frac{d(x^2-5x+8)}{dx}+(x^2-5x+8)\frac{d(x^3+7x+9)}{dx}$

$\Rightarrow \frac{dy}{dx}=(x^3+7x+9)(2x-5)+(x^2-5x+8)(3x^2+7)$

$\Rightarrow \frac{dy}{dx}=2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56$

$\therefore \frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11$

(ii) Let $y=(x^2-5x+8)(x^3+7x+9)$

Expanding the product to obtain a single polynomial

$y=(x^2-5x+8)(x^3+7x+9)$
$y=x^2(x^3+7x+9)-5x(x^3+7x+9)+8(x^3+7x+9)$

$y=x^5+7x^3+9x^2-5x^4-35x^2-45x+8x^3+56x+72$

$y=x^5-5x^4+15x^3-26x^2+11x+72$

Differentiating both sides w.r.t. $x$ , we get,

$\frac{dy}{dx}=\frac{d(x^5-5x^4+15x^3-26x^2+11x+72)}{dx}$

$\frac{dy}{dx}=\frac{d\left(x^5\right)}{dx}-\frac{d\left(5x^4\right)}{dx}+\frac{d\left(15x^3\right)}{dx}-\frac{d\left(26x^2\right)}{dx}+\frac{d\left(11x\right)}{dx}+\frac{d\left(72\right)}{dx}$

$\frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11+0$

$\frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11$

(iii) Let $y=(x^2-5x+8)(x^3+7x+9)$

Taking $\text{log}$ both sides, we get,
$\text{log}y=\text{log}\left(\right(x^2-5x+8\left)\right(x^3+7x+9\left)\right)$

Using $\text{log}ab=\text{log}a+\text{log}b$

$\text{log}y=\text{log}(x^2-5x+8)+\text{log}(x^3+7x+9)$

Differentiating both sides w.r.t. $x$ , we get,

$\frac{d\left(\text{log}y\right)}{dx}=\frac{d\left(\text{log}\right(x^2-5x+8)+\text{log}(x^3+7x+9\left)\right)}{dx}$

$\frac{d\left(\text{log}y\right)}{dy}.\frac{dy}{dx}=\frac{d\left(\text{log}\right(x^2-5x+8\left)\right)}{dx}+\frac{d\left(\text{log}\right(x^3+7x+9\left)\right)}{dx}$

$\frac{1}{y}.\frac{dy}{dx}=\frac{1}{(x^2-5x+8)}.\frac{d\left(\text{log}\right(x^2-5x+8\left)\right)}{dx}+\frac{1}{(x^3+7x+9)}.\frac{d\left(\text{log}\right(x^3+7x+9\left)\right)}{dx}$

$\frac{1}{y}.\frac{dy}{dx}=\frac{2x-5}{x^2-5x+8}+\frac{3x^2+7}{x^3+7x+9}$

$\frac{1}{y}.\frac{dy}{dx}=\frac{\left(\right(2x-5\left)\right(x^3+7x+9)+(3x^2+7\left)\right(x^2-5x+8\left)\right)}{(x^2-5x+8)(x^3+7x+9)}$

$\frac{dy}{dx}=y\left(\frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)}\right)$

Putting $y=(x^2-5x+8)(x^3+7x+9)$, we get

$\frac{dy}{dx}=(x^2-5x+8)(x^3+7x+9)\left(\frac{(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)}{(x^2-5x+8)(x^3+7x+9)}\right)$

$\frac{dy}{dx}=(2x-5)(x^3+7x+9)+(3x^2+7)(x^2-5x+8)$

$\frac{dy}{dx}=2x(x^3+7x+9)-5(x^3+7x+9)+3x^2(x^2-5x+8)+7(x^2-5x+8)$

$\frac{dy}{dx}=2x^4+14x^2+18x-5x^3-35x-45+3x^4-15x^3+24x^2+7x^2-35x+56$

$\frac{dy}{dx}=5x^4-20x^3+45x^2-52x+11$