Question

differentiate x^8/8^x with respect to X​

Answer 1

Answer:

8^x8x^7-x^8 8log8^x/8^x2

Step-by-step explanation:

use quotient rule to solve this

Answer 2

Answer:

$\large \boxed{\sf{ \dfrac{8 {x}^{7} - {x}^{8} ln(8) }{ {8}^{x} } }}$

Step-by-step explanation:

$y = \dfrac{ {x}^{8} }{ {8}^{x} }$

We know that,

$\dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{ g(x) \dfrac{d}{dx} f(x) - f(x) \dfrac{d}{dx} g(x)}{ {(g(x))}^{2} }$

Therefore, differentiate wrt x , we get

$= > \dfrac{dy}{dx} = \dfrac{ {8}^{x} \dfrac{d}{dx} {x}^{8} - {x}^{8} \dfrac{d}{dx} {8}^{x} }{ {( {8}^{x} )}^{2} }$

But, we know that,

• $\dfrac{d}{dx} {a}^{x} = {a}^{x} ln(a)$

• $\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}$

Therefore, we will get,

$= > \dfrac{dy}{dx} = \dfrac{ {8}^{x} \times 8 {x}^{7} - {x}^{8} \times {8}^{x} ln(8) }{ {8}^{2x} } \\ \\ = > \dfrac{dy}{dx} = \dfrac{ \cancel{{8}^{x}}(8 {x}^{7} - {x}^{8} ln8) }{ \cancel{ {8}^{x} } \times {8}^{x} } \\ \\ = > \large \bold{ \dfrac{dy}{dx} = \dfrac{8 {x}^{7} - {x}^{8} ln(8) }{ {8}^{x} } }$