If y = (1 + x)^x,then frac{dy}{dx} = | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given $y = (1 + x)^x$. Taking the natural logarithm on both sides,we get $\log y = x \log(1 + x)$. Differentiating both sides with respect to $x$

Vedclass · Accepted Answer

$(1 + x)^x \left[ \frac{x}{1 + x} + \log(1 + x) \right]$

Vedclass · Answer

(C) Given $y = (1 + x)^x$. Taking the natural logarithm on both sides,we get $\log y = x \log(1 + x)$. Differentiating both sides with respect to $x$ using the product rule: $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[x] \cdot \log(1 + x) + x \cdot \frac{d}{dx}[\log(1 + x)]$. $\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log(1 + x) + x \cdot \frac{1}{1 + x}$. Multiplying both sides by $y$,we get $\frac{dy}{dx} = y \left[ \log(1 + x) + \frac{x}{1 + x} \right]$. Substituting $y = (1 + x)^x$,we obtain $\frac{dy}{dx} = (1 + x)^x \left[ \frac{x}{1 + x} + \log(1 + x) \right]$.

If $y = (1 + x)^x$,then $\frac{dy}{dx} = $

Similar Questions

$\text{If } \frac{d}{dx} \left( \frac{x^2+1}{(x^2+5)(x^2+9)} \right) = \frac{2x(x^2+1)}{(x^2+5)(x^2+9)} \left[ \frac{1}{f(x)} - \frac{1}{g(x)} - \frac{1}{h(x)} \right], \text{ then } 2h(x) - f(x) - g(x) = $

If $y = \sqrt {\frac{(x - a)(x - b)}{(x - c)(x - d)}} $,then $\frac{dy}{dx} = $

If $y=x^{\sqrt{x}}$,then $\frac{dy}{dx}=$

If $f(x) = \frac{e^{-x} \sin x}{\log_e x}$ and $f'(x) = f(x) \cdot g(x)$,then $g'(e) =$

The derivative of $y = x^{\left(x^x\right)}$ with respect to $x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module