For constant a,frac{d}{d x}left(x^{x}+x^{a}+a | vedclass

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

Question

(B) We need to find the derivative of the sum: $\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$.Using the linearity property of the derivative,we h

Vedclass · Accepted Answer

$x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$

Vedclass · Answer

(B) We need to find the derivative of the sum: $\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$.Using the linearity property of the derivative,we have:$\frac{d}{d x}\left(x^{x}\right)+\frac{d}{d x}\left(x^{a}\right)+\frac{d}{d x}\left(a^{x}\right)+\frac{d}{d x}\left(a^{a}\right)$.$1$. For $\frac{d}{d x}(x^x)$: Let $y = x^x$. Taking $\log$ on both sides,$\log y = x \log x$. Differentiating with respect to $x$,$\frac{1}{y} \frac{dy}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{dy}{dx} = x^x(1 + \log x)$.$2$. For $\frac{d}{d x}(x^a)$: Using the power rule,$\frac{d}{d x}(x^a) = a x^{a-1}$.$3$. For $\frac{d}{d x}(a^x)$: Using the exponential derivative rule,$\frac{d}{d x}(a^x) = a^x \log a$.$4$. For $\frac{d}{d x}(a^a)$: Since $a$ is a constant,$a^a$ is a constant,so its derivative is $0$.Combining these,the result is $x^{x}(1+\log x)+a x^{a-1}+a^{x} \log a$.

For constant $a$,$\frac{d}{d x}\left(x^{x}+x^{a}+a^{x}+a^{a}\right)$ is

Similar Questions

If $f(x) = \frac{x}{1 + |x|}$ for $x \in R$,then $f'(0) = $

Find the derivative of the following function (it is to be understood that $a, b, c, d, p, q, r$ and $s$ are fixed non-zero constants and $m$ and $n$ are integers): $\frac{a+b \sin x}{c+d \cos x}$

$\frac{d}{dx}\left[ \frac{2}{\pi }\sin {x^\circ} \right] = $

If $y=\tan ^{-1}(\sin \sqrt{x})+\operatorname{cosec}^{-1}\left(e^{2 x+1}\right)$,then $\frac{d y}{d x}=$

The differential coefficient of $\sqrt{\sec \sqrt{x}}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module