Question

If y 1/2x+3then dn y dx n..

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{y=\dfrac{1}{2x+3}}$

Differentiating w.r.t. x, successively,

$\tt{y_{1}=(2)\dfrac{(-1)}{(2x+3)^2}}$

$\tt{\implies\,y_{2}=(2)^2\dfrac{(-1)(-2)}{(2x+3)^3}}$

$\tt{\implies\,y_{3}=(2)^3\dfrac{(-1)(-2)(-3)}{(2x+3)^4}}$

$\tt{\implies\,y_{4}=(2)^4\dfrac{(-1)(-2)(-3)(-4)}{(2x+3)^5}}$

$\tt{\implies\,y_{n}=(2)^n\dfrac{(-1)^n\,n!}{(2x+3)^{n+1}}}$

$\tt{\implies\,y_{n}=\dfrac{(-1)^n\cdot\,(2)^n\cdot\,n!}{(2x+3)^{n+1}}}$