Question

the value of the term lying between first and eleventh term of an arithmetic progression is 35 find the value of 11 and 65​

Answer 1

$\begin{gathered} \orange{ \boxed{\boxed{\begin{array}{cc} \bf \to \: Let, \\ \\ \rm \: y = ln(1 + x) \\ \\ \rm \implies \: \frac{dy}{dx} = \frac{d}{dx} \: \{ ln(1 + x) \} \\ \\ \pink{ {\boxed{\begin{array}{cc} \sf \: we \: know \: that : \\ \\ \rm \frac{d}{dx} \: ln \: x = \frac{1}{x} \\ \\ \rm \frac{d}{dx} \: {x}^{n} = n {x}^{n - 1} \\ \\ \rm \: \frac{d}{dx} (constant) = 0 \end{array}}}} \\ \: \: \sf \: apply \: this \: \\ \\ \rm = \frac{1}{1 + x}. \frac{d}{dx} (1 + x) \\ \\ \rm = \frac{1}{1 + x} \{ \frac{d}{dx} \: 1 + \frac{d}{dx} x \} \\ \\ \rm = \frac{1}{1 + x} (0 + 1) \\ \\ \rm = \frac{1}{1 + x} \\ \\ \\ \blue{ \boxed{ \therefore \rm \: \frac{d}{dx} \: ln(1 + x) = \frac{1}{1 + x}}} \end{array}}}}\end{gathered}$