Find $${d}^{2}y/d{x}^{2}$$

$$y=x\mathrm{sin}x-3\mathrm{cos}x$$

$$y=x\mathrm{sin}x-3\mathrm{cos}x$$

Ryder Ferguson
2022-10-22
Answered

Find $${d}^{2}y/d{x}^{2}$$

$$y=x\mathrm{sin}x-3\mathrm{cos}x$$

$$y=x\mathrm{sin}x-3\mathrm{cos}x$$

You can still ask an expert for help

Gael Irwin

Answered 2022-10-23
Author has **13** answers

Given that,

$$y=x\mathrm{sin}x-3\mathrm{cos}x$$

Differentiate w.r.t x

$$\frac{dy}{dx}=x(\mathrm{cos}x)+\mathrm{sin}x(1)-(-3\mathrm{sin}x)\phantom{\rule{0ex}{0ex}}=x\mathrm{cos}x+\mathrm{sin}x+3\mathrm{sin}x\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=x\mathrm{cos}x+4\mathrm{sin}x$$

Again Diff. w.r.t. x

$$\frac{{d}^{2}y}{d{x}^{2}}=x\cdot (-\mathrm{sin}x)+\mathrm{cos}x(1)+4\mathrm{cos}x\phantom{\rule{0ex}{0ex}}=-x\mathrm{sin}x+\mathrm{cos}x+4\mathrm{cos}x\phantom{\rule{0ex}{0ex}}\frac{{d}^{2}y}{d{x}^{2}}=-x\mathrm{sin}x+5\mathrm{cos}x$$

$$y=x\mathrm{sin}x-3\mathrm{cos}x$$

Differentiate w.r.t x

$$\frac{dy}{dx}=x(\mathrm{cos}x)+\mathrm{sin}x(1)-(-3\mathrm{sin}x)\phantom{\rule{0ex}{0ex}}=x\mathrm{cos}x+\mathrm{sin}x+3\mathrm{sin}x\phantom{\rule{0ex}{0ex}}\frac{dy}{dx}=x\mathrm{cos}x+4\mathrm{sin}x$$

Again Diff. w.r.t. x

$$\frac{{d}^{2}y}{d{x}^{2}}=x\cdot (-\mathrm{sin}x)+\mathrm{cos}x(1)+4\mathrm{cos}x\phantom{\rule{0ex}{0ex}}=-x\mathrm{sin}x+\mathrm{cos}x+4\mathrm{cos}x\phantom{\rule{0ex}{0ex}}\frac{{d}^{2}y}{d{x}^{2}}=-x\mathrm{sin}x+5\mathrm{cos}x$$

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When I solve this equation my answer comes to be $x=3\pi /4\pm 2n\pi ,15\pi /4\pm 2n\pi $ where $n$ is an integer

However when I graph the equation $y=\mathrm{tan}(x/3)-1$ values for $x$ such as $11\pi /4$ do not equal zero.

What would x be equal to then?

Solve for general solutions

$\mathrm{tan}(x/3)=1$

When I solve this equation my answer comes to be $x=3\pi /4\pm 2n\pi ,15\pi /4\pm 2n\pi $ where $n$ is an integer

However when I graph the equation $y=\mathrm{tan}(x/3)-1$ values for $x$ such as $11\pi /4$ do not equal zero.

What would x be equal to then?

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Show that,

$2\mathrm{arctan}\left(\frac{1}{3}\right)+2\mathrm{arcsin}\left(\frac{1}{5\sqrt{2}}\right)-\mathrm{arctan}\left(\frac{1}{7}\right)=\frac{\pi}{4}$

There is a mixed of sin and tan, how can I simplify this to $\frac{\pi}{4}$

We know the identity of $\mathrm{arctan}\left(\frac{1}{a}\right)+\mathrm{arctan}\left(\frac{1}{b}\right)=\mathrm{arctan}\left(\frac{a+b}{ab-1}\right)$

$2\mathrm{arctan}\left(\frac{1}{3}\right)+2\mathrm{arcsin}\left(\frac{1}{5\sqrt{2}}\right)-\mathrm{arctan}\left(\frac{1}{7}\right)=\frac{\pi}{4}$

There is a mixed of sin and tan, how can I simplify this to $\frac{\pi}{4}$

We know the identity of $\mathrm{arctan}\left(\frac{1}{a}\right)+\mathrm{arctan}\left(\frac{1}{b}\right)=\mathrm{arctan}\left(\frac{a+b}{ab-1}\right)$