Question

Find the derivative of x/x+1 wrt w​

Answer 1

Answer :-

Differentiating by using Quotient rule -

$\implies\sf\dfrac{d}{dx}\Big[\dfrac{u}{v}\Big] = \dfrac{ v\dfrac{du}{dx} - u \dfrac{dv}{dx} }{v^2}$

For the given question :-

• u = x
• v = x + 1

$\implies\sf\dfrac{\dfrac{d}{dx}\Big[\dfrac{x}{x +1}\Big] = ( x + 1 )\Big[\dfrac{d}{dx}(x)\Big] - x \Big[\dfrac{d}{dx}(x + 1)\Big]}{(x + 1)^2}$

ㅤ

• $\sf \dfrac{d(x)}{dx}= 1$

ㅤ

• $\sf \dfrac{d(x+1)}{dx} = \dfrac{d(x)}{dx} + \dfrac{d(1)}{dx} = 1 + 0 = 1$

ㅤ

$\implies\sf\dfrac{( x + 1 )1 - x ( 1 )}{(x^2 + 1 + 2x)}$

$\implies\sf\dfrac{1 + \cancel x - \cancel x}{x^2 + 1 + 2x}$

$\implies\sf\dfrac{1}{x^2 + 1 + 2x}$

$\boxed{\sf\dfrac{d}{dx}\Big[\dfrac{x}{x+1}\Big] = \dfrac{1}{x^2 + 1 + 2x}}$

Additional information :-

Derivative of constant -

$\implies\sf\dfrac{d}{dx}(c) = 0$

Power rule -

$\implies\sf\dfrac{d}{dx}x^n = nx^{n-1}$

Sum rule -

$\implies\sf\dfrac{d}{dx}(u + v) = \dfrac{du}{dx} + \dfrac{dv}{dx}$

Product Rule -

$\implies\sf\dfrac{d}{dx}(uv) = u\dfrac{dv}{dx} + v\dfrac{du}{dx}$

$\boxed{\boxed{\begin{minipage}{5cm}\displaystyle\circ\sf\;\dfrac{d}{dx}(sin\;x)=cosx \\\\ \circ \;\dfrac{d}{dx}(cos\;x) = -sinx \\\\ \circ \; \dfrac{d}{dx}(tan\;x) = sec^{2}x \\\\ \circ\; \dfrac{d}{dx}(cot\;x) = -csc^{2}x \\\\ \circ \; \dfrac{d}{dx}(sec\;x) = secx \cdot tanx \\\\ \circ \; \dfrac{d}{dx}(csc\;x) = -cscx \cdot cotx \\\\ \circ\; \dfrac{d}{dx}(sinh\;x)=coshx \\\\ \circ\; \dfrac{d}{dx}(cosh\;x)= sinhx \\\\ \circ\;\dfrac{d}{dx}(tanh\;x)=sech^{2}h \\\\ \circ\;\dfrac{d}{dx}(coth\;x)=-csch^{2}x \\\\ \circ\;\dfrac{d}{dx}(sech\;x) =-sechx \cdot tanhx \\\\ \circ\;\dfrac{d}{dx}(csch\;x) = -cschx \cdot cothx\end{minipage}}}$