Question

find dy/dx of f(x)= (2-x)^6 (5+2x)^4​

Answer 1

Answer: df/dx = (x-2)^5 ( 2x + 5 )^3 (20x + 14)

Explanation:

For finding the derivative of product of functions of the independent variable one can use the product rule of differentiation. Which is as follows

For a function f(x) = u(x) v(x) ⇒ df/dx = u(x) dv/dx + v(x) du/dx

For the given function one can take u(x) = (2 - x)^6 and v(x) = (5 + 2x)^4

du/dx = -6 (2-x)^5

dv/dx = 8 (5+2x)^3

Simplifying we get df/dx = (x-2)^5 ( 2x + 5 )^3 (20x + 14)

Answer 2

Need to FinD :-

• The derivative of the given function .

$\red{\frak{Given}}\Bigg\{ \sf f(x) = (2-x)^6 (5+2x)^4$

Here we can use the product rule of differenciation . Say we have two functions u and v and we need to differentiate uv . Then its differenciation with respect to x , will be , Let y = f(x) = uv ,

$\twoheadrightarrow\boxed{\sf \red{\dfrac{dy}{dx}= u\dfrac{dv}{dx}+v\dfrac{du}{dx} }}$

Also here u and v have some Power . So we will also use Power Chain Rule . Say if we have a function U and the power of function is n , then its derivative will be , ( y = Uⁿ)

$\twoheadrightarrow \boxed{\sf \red{ \dfrac{dy}{dx}= nU^{n-1}\dfrac{dU}{dx} }}$

Now here ,

• u = (2 - x)⁶
• v = (5 + 2x)⁴

• So that ,

$\dashrightarrow \sf \dfrac{dy}{dx}= \dfrac{d}{dy}( 2-6)^6 (5+2x)^4\\\\\sf\dashrightarrow \dfrac{dy}{dx} = (2-x)^6\bigg[ \dfrac{d}{dx}(2x+5)^4 \bigg] + (2x+5)^4\bigg[ \dfrac{d}{dx} (2-x)^6 \bigg] \\\\\sf\dashrightarrow \dfrac{dy}{dx} = (2-x)^6\bigg[ \dfrac{d(2x+5)}{dx}4(2x+5)^3\bigg] + (2x+5)^4\bigg[ \dfrac{d(2-x)}{dx}6(2-x)^5\bigg] \\\\\sf\dashrightarrow \dfrac{dy}{dx} = (2-x)^6[ 2.4(2x+5)^3 ] + (2x+5)^4[ (-1).6(2-x)^5] \\\\\sf\dashrightarrow \dfrac{dy}{dx} = 8(2-x)^6(2x+5) -6(2x+5)^4(2-x)^5 \\\\\sf\dashrightarrow \dfrac{dy}{dx} = 2 (2-x)^5(2x+5)^3[ 4(2-x) -3(2x+5)] \\\\\sf\dashrightarrow \dfrac{dy}{dx} = 2(2-x)^5(2x+5)^3[ 8 - 4x - 6x -15 ] \\\\\sf\dashrightarrow \dfrac{dy}{dx} = 2(2-x)^5(2x+5)^3 [ -10x - 7 ] \\\\\sf\dashrightarrow \underset{\blue{\sf Required\: Derivative}}{\underbrace{\boxed{\pink{\frak{ \dfrac{dy}{dx} = f'(x) = -2(2-x)^5(2x+5)^3(10x+7) }}}}}$