If y = frac{sqrt[3]{1 + 3x} sqrt[4]{1 + 4x} s | vedclass

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

Question

(B) Given $y = \frac{(1 + 3x)^{1/3} (1 + 4x)^{1/4} (1 + 5x)^{1/5}}{(1 + 7x)^{1/7} (1 + 8x)^{1/8}}$.Taking the natural logarithm on both sides:$\ln y =

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) Given $y = \frac{(1 + 3x)^{1/3} (1 + 4x)^{1/4} (1 + 5x)^{1/5}}{(1 + 7x)^{1/7} (1 + 8x)^{1/8}}$.Taking the natural logarithm on both sides:$\ln y = \frac{1}{3} \ln(1 + 3x) + \frac{1}{4} \ln(1 + 4x) + \frac{1}{5} \ln(1 + 5x) - \frac{1}{7} \ln(1 + 7x) - \frac{1}{8} \ln(1 + 8x)$.Differentiating with respect to $x$:$\frac{1}{y} \frac{dy}{dx} = \frac{1}{3} \cdot \frac{3}{1 + 3x} + \frac{1}{4} \cdot \frac{4}{1 + 4x} + \frac{1}{5} \cdot \frac{5}{1 + 5x} - \frac{1}{7} \cdot \frac{7}{1 + 7x} - \frac{1}{8} \cdot \frac{8}{1 + 8x}$.$\frac{1}{y} \frac{dy}{dx} = \frac{1}{1 + 3x} + \frac{1}{1 + 4x} + \frac{1}{1 + 5x} - \frac{1}{1 + 7x} - \frac{1}{1 + 8x}$.At $x = 0$,note that $y(0) = \frac{\sqrt[3]{1} \sqrt[4]{1} \sqrt[5]{1}}{\sqrt[7]{1} \sqrt[8]{1}} = 1$.Substituting $x = 0$ into the derivative expression:$\frac{1}{1} \cdot y'(0) = (1 + 1 + 1 - 1 - 1) = 1$.Therefore,$y'(0) = 1$.

If $y = \frac{\sqrt[3]{1 + 3x} \sqrt[4]{1 + 4x} \sqrt[5]{1 + 5x}}{\sqrt[7]{1 + 7x} \sqrt[8]{1 + 8x}}$,then $y'(0)$ is equal to

Similar Questions

If $y = e^{4x} \left( \frac{x-4}{x+3} \right)^{\frac{3}{4}}$,then $\frac{dy}{dx} = $

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

If $x^{m} y^{n}=(x+y)^{m+n}$,then $\frac{dy}{dx}$ is equal to

If $y = \log \left[e^{5x} \left(\frac{3x-4}{x+5}\right)^{\frac{4}{3}}\right]$,then $\frac{dy}{dx}$ is equal to

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module