Question

Differentiate the functions in, (x + 3)².(x + 4)³ .(x + 5)⁴

Answer 1

Step-by-step explanation:

$\sf {\underline{\bigstar Let\ y\ =\ (x\ +\ 3)^2\ .(x\ +\ 4)^3\ .(x\ +\ 5)^4}}$

➽ Taking log both sides :-

$\sf \mapsto {log\ y\ =\ log \big((x\ +\ 3)^2\ .(x\ +\ 4)^3\ .(x\ +\ 5)^4 \big)}$

$\sf \mapsto {log\ y\ =\ log(x\ +\ 3)^2\ +\ log(x\ +\ 4)^3\ +\ log(x\ +\ 5)^4}$

$\sf \mapsto {log\ y\ =\ 2\ log(x\ +\ 3)\ +\ 3\ log(x\ +\ 4)\ +\ 4\ log(x\ +\ 5)}$

$\sf {\underline{\bigstar Differentiating\ both\ sides\ w.r.t.x.}}$

$\sf : \implies {\dfrac{d(log\ y)}{dx}\ =\ \dfrac{d \big(2\ log(x\ +\ 3)\ +\ 3\ log(x\ +\ 4)\ +\ 4\ log(x\ +\ 5) \big)}{dx}}$

$\sf : \implies {\dfrac{d(log\ y)}{dx} \bigg(\dfrac{dy}{dy} \bigg)\ =\ \dfrac{d \big(2\ log(x\ +\ 3) \big)}{dx}\ +\ \dfrac{d \big(3\ log(x\ +\ 4) \big)}{dx}\ +\ \dfrac{d \big(4\ log(x\ +\ 5) \big)}{dx}}$

$\sf : \implies {\dfrac{d(log\ y)}{dy} \bigg(\dfrac{dy}{dx} \bigg)\ =\ 2\ \dfrac{d \big(log(x\ +\ 3) \big)}{dx}\ +\ 3\ \dfrac{d \big(log(x\ +\ 4) \big)}{dx}\ +\ 4\ \dfrac{d \big(log(x\ +\ 5) \big)}{dx}}$

$\sf : \implies {\dfrac{1}{y} \times \dfrac{dy}{dx}\ =\ 2.\ \dfrac{1}{(x\ +\ 3)}.\ \dfrac{d(x\ +\ 3)}{dx}\ +\ 3.\ \dfrac{1}{(x\ +\ 4)}.\ \dfrac{d(x\ +\ 4)}{dx}\ +\ 4.\ \dfrac{1}{(x\ +\ 5)}.\ \dfrac{d(x\ +\ 5)}{dx}}$

$\sf : \implies {\dfrac{1}{y} \times \dfrac{dy}{dx}\ =\ \dfrac{2}{(x\ +\ 3)} \bigg(\dfrac{dx}{dx}\ +\ \dfrac{d(3)}{dx} \bigg)\ +\ \dfrac{3}{x\ +\ 4} \bigg(\dfrac{dx}{dx}\ +\ \dfrac{d(4)}{dx} \bigg)\ +\ \dfrac{4}{x\ +\ 5} \bigg(\dfrac{dx}{dx}\ +\ \dfrac{d(5)}{dx} \bigg)}$

$\sf : \implies {\dfrac{1}{y} \times \dfrac{dy}{dx}\ =\ \dfrac{2}{x\ +\ 3}\ (1\ +\ 0)\ +\ \dfrac{3}{x\ +\ 4}\ (1\ +\ 0)\ +\ \dfrac{4}{x\ +\ 5}\ (1\ +\ 0)}$

$\sf : \implies {\dfrac{1}{y} \times \dfrac{dy}{dx}\ =\ \dfrac{2}{x\ +\ 3}\ +\ \dfrac{3}{x\ +\ 4}\ +\ \dfrac{4}{x\ +\ 5}}$

$\sf : \implies {\dfrac{dy}{dx}\ =\ y \bigg(\dfrac{2}{x\ +\ 3}\ +\ \dfrac{3}{x\ +\ 4}\ +\ \dfrac{4}{x\ +\ 5} \bigg)}$

➽ By putting the value of y :-

$\sf : \implies {\dfrac{dy}{dx}\ =\ (x\ +\ 3)^2\ .(x\ +\ 4)^3\ .(x\ +\ 5)^4 \bigg(\dfrac{2}{(x\ +\ 3)}\ +\ \dfrac{3}{(x\ +\ 4)}\ +\ \dfrac{4}{(x\ +\ 5)} \bigg)}$

$\sf : \implies {\dfrac{dy}{dx}\ =\ (x\ +\ 3)^2\ (x\ +\ 4)^3\ (x\ +\ 5)^4 \bigg( \dfrac{2(x\ +\ 4)(x\ +\ 5)\ +\ 3(x\ +\ 3)(x\ +\ 5)\ +\ 4(x\ +\ 3)(x\ +\ 4)}{(x\ +\ 3)(x\ +\ 4)(x\ +\ 5)} \bigg)}$

$\sf : \implies {\dfrac{dy}{dx}\ =\ \dfrac{(x\ +\ 3)^2(x\ +\ 4)^3(x\ +\ 5)^4}{(x\ +\ 3)(x\ +\ 4)(x\ +\ 5)}}$

$\sf \: \: \: \: \: \: \: \: {=\ \big(2(x^2\ +\ 4x\ +\ 5x\ +\ 20)\ +\ 3(x^2\ +\ 3x\ +\ 5x\ +\ 15)\ +\ 4(x^2\ +\ 3x\ +\ 4x\ +\ 12) \big)}$

$\sf : \implies {\dfrac{dy}{dx}\ =\ (x\ +\ 3)(x\ +\ 4)^2(x\ +\ 5)^3}$

$\sf \: \: \: \: \: \: \: \: {=\ \big(2(x^2\ +\ 9x\ +\ 20)\ +\ 3(x^2\ +\ 8x\ +\ 15)\ +\ 4(x^2\ +\ 7x\ +\ 12) \big)}$

$\sf : \implies {\dfrac{dy}{dx}\ =\ (x\ +\ 3)(x\ +\ 4)^2(x\ +\ 5)^3}$

$\sf \: \: \: \: \: \: \: \: {=\ (2x^2\ +\ 18x\ +\ 40\ +\ 3x^2\ +\ 24x\ +\ 45\ +\ 4x^2\ +\ 28x\ +\ 48)}$

$\sf : \implies {\dfrac{dy}{dx}\ =\ (x\ +\ 3)(x\ +\ 4)^2(x\ +\ 5)^3}$

$\sf \: \: \: \: \: \: \: \: {=\ (2x^2\ +\ 3x^2\ +\ 4x^2 18x\ +\ 24x\ +\ 28x\ +\ 40\ +\ 45\ +\ 48)}$

$\sf \therefore {\underline{\boxed{\sf \dfrac{dy}{dx}\ =\ (x\ +\ 3)(x\ +\ 4)^2(x\ +\ 5)^3(9x^2\ +\ 70x\ +\ 133).}}}$