If y=x^x+x^7+7^x+7^7,then frac{dy}{dx}= | vedclass

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

Question

(B) Given $y = x^x + x^7 + 7^x + 7^7$. To find $\frac{dy}{dx}$,we differentiate each term with respect to $x$ separately. $1$. For $x^x$: Let $u = x^x

Vedclass · Accepted Answer

$x^x(1+\log_e x)+7 x^6+7^x(\log_e 7)$

Vedclass · Answer

(B) Given $y = x^x + x^7 + 7^x + 7^7$. To find $\frac{dy}{dx}$,we differentiate each term with respect to $x$ separately. $1$. For $x^x$: Let $u = x^x$. Taking $\log$ on both sides,$\log u = x \log x$. Differentiating with respect to $x$,$\frac{1}{u} \frac{du}{dx} = 1 \cdot \log x + x \cdot \frac{1}{x} = \log x + 1$. Thus,$\frac{du}{dx} = x^x(1 + \log_e x)$. $2$. For $x^7$: Using the power rule,$\frac{d}{dx}(x^7) = 7x^6$. $3$. For $7^x$: Using the exponential rule $\frac{d}{dx}(a^x) = a^x \log_e a$,we get $\frac{d}{dx}(7^x) = 7^x \log_e 7$. $4$. For $7^7$: Since $7^7$ is a constant,its derivative is $0$. Combining these,$\frac{dy}{dx} = x^x(1 + \log_e x) + 7x^6 + 7^x \log_e 7 + 0$. Therefore,$\frac{dy}{dx} = x^x(1 + \log_e x) + 7x^6 + 7^x \log_e 7$.

If $y=x^x+x^7+7^x+7^7$,then $\frac{dy}{dx}=$

Similar Questions

Find the derivative of $(5x^{3} + 3x - 1)(x - 1)$.

Find the derivative: $\frac{d}{dx}\{ \cos(\sin(x^2)) \}$

If $f(x) = \sqrt{ax} + \frac{a^2}{\sqrt{ax}}$,then $f^{\prime}(a)$ is equal to

If $y = e^{\sqrt{x}} + e^{-\sqrt{x}}$,then $\frac{dy}{dx}$ equals:

If $y=\left(\log _{\cot x} \tan x\right)\left(\log _{\tan x} \cot x\right)+\tan ^{-1}\left(\frac{4 x}{4-x^2}\right)$,then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module