Question

$\displaystyle \tt \frac{ {d}^{ - 1} }{d {x}^{ - 1} } \bigg( \frac{\pi e}{x( \varphi + lnx) {}^{ \sqrt{2} } } \bigg)$​

Answer 1

Answer:

$\sf\fbox\red{Answer:-}$

d (1+x² + x¹ dx1+x+x²S = ax + b d f1+2x² + x¹ - x² dx 1+x+x² = ax + b

d (1+x²)² - x² dx 1+x+x² = ax + b

d dx 介 -x}} + x² + x) (1 + x² − x) 1+x+x² = ax + b

d

dx (1 + x² − x) ⇒ 2x1 = ax + b

⇒a=2, b = - 1

>

.. (a, b) = (2, - 1)

Answer 2

Answer:

$\quad \displaystyle \bold \green{ \frac{e \pi}{ {d}^{2}( \varphi + ln(x)^{ \sqrt{2}} } }$

Step-by-step explanation:

$\quad \displaystyle \bold \red{ \frac{d^{ - 1} }{dx^{ - 1} } \frac{ \pi \: e}{x( \varphi + ln(x)) {}^{ \sqrt{2} } }}$

$\quad \displaystyle \bold \red{ \frac{x}{ {d}^{2} } \frac{e \pi}{x( \varphi + ln(x) {}^{ \sqrt{2} } } }$

$\quad \displaystyle \bold \red{ = \frac{e \pi}{ {d}^{2}( \varphi + ln(x) {}^{ \sqrt{2} } }}$