Best solutions to - "Find the derivative y=ln⁡xx6 y=ex5ln⁡x Find the indicated..."

Elleanor Mckenzie

Answered question

2021-05-13

Find the derivative

y = \frac{\ln x}{x^{6}}

y = e^{x^{5}} \ln x

Find the indicated derivative of the function.

\frac{d^{4} y}{d x^{4}} of y = 2 x^{5} - 3 x^{2} - 5 x + 1

pivonie8

Skilled2021-05-14Added 91 answers

The solution is given below:

Jeffrey Jordon

Expert2021-09-30Added 2605 answers

1) $y = \frac{\ln x}{x^{6}}$

it is of the form $\frac{4}{v}$ so $d \frac{u}{v} = \frac{v u - u v}{v^{2}}$

$\Rightarrow \frac{x^{6} \frac{d}{d x} (\ln x) - \ln x \frac{d}{d x} (x^{6})}{(x^{6})^{2}}$

$\Rightarrow \frac{x^{x^{5}} \frac{1}{x} - \ln x \cdot 6 x^{5}}{x^{12}}$

$\frac{x^{5} - 6 \ln x x^{5}}{x^{12}} \Rightarrow \frac{x^{5} (1 - 6 \ln x)}{x^{12}} \Rightarrow \frac{d y}{d x} = \frac{1 - 6 \ln x}{x^{7}}$

2) $y = e^{x^{5}} \ln x$

$\frac{d}{d x} (u v) = u v + v u$

$\Rightarrow e^{x^{5}} \frac{d}{d x} (\ln x) + \ln x \frac{d}{d x} (e^{x^{5}})$

$= e^{x^{5}} \frac{1}{x} + \ln x \cdot e^{x^{5}} \frac{d}{d x} (x^{5})$

$= \frac{e^{x^{5}}}{x} + \ln x e^{x^{5}} 5 x^{4}$

$\frac{d y}{d n} = e^{x^{5}} [\frac{1}{x} + 5 \ln x x^{4}]$

Given $y = 2 x^{5} - 3 x^{2} - 5 x + 1$

$\frac{d y}{d x} = 2 (5 x^{5 - 1}) - 3 (2 \cdot x^{2 - x}) - 5 (1) + 0$

$\frac{d}{d x} (x^{4}) = n \cdot x^{n - 1}$

$\frac{d y}{d x} \Rightarrow 10 x^{4} - 6 x - 5$

Again derivative w.r to x

$\frac{d^{2} y}{d x^{2}} = 10 (4 x^{3}) - 6 (1) - 0$

$\frac{d^{2} y}{d x^{2}} = 40 x^{3} - 6$

Again derivative w. r. to x

$\frac{d^{3} y}{d x^{3}} = 40 (3 x^{2}) - 0$

$\frac{d^{3} y}{d x^{3}} = 120 x^{2}$

Again derivative w.r. to x

$\frac{d^{4} x}{2 x^{4}} = 120 (2 x) = 240 x$

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