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2 years ago

Use implicit differentiation to solve that the derivative

Mathematics

1 answer:

Len [333]2 years ago

7 0

Given

e ˣʸ = sec(x ²)

take the derivative of both sides:

d/dx [e ˣʸ] = d/dx [sec(x ²)]

Use the chain rule:

e ˣʸ d/dx [xy] = sec(x ²) tan(x ²) d/dx [x ²]

Use the product rule on the left, and the power rule on the right:

e ˣʸ (x dy/dx + y) = sec(x ²) tan(x ²) (2x)

Solve for dy/dx :

e ˣʸ (x dy/dx + y) = 2x sec(x ²) tan(x ²)

x dy/dx + y = 2x e ⁻ˣʸ sec(x ²) tan(x ²)

x dy/dx = 2x e ⁻ˣʸ sec(x ²) tan(x ²) - y

dy/dx = 2e ⁻ˣʸ sec(x ²) tan(x ²) - y/x

Since e ˣʸ = sec(x ²), we simplify further to get

dy/dx = 2 tan(x ²) - y/x

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(I tried to answer your question - even though you posted no question - AND you posted no graphics)

Step-by-step explanation:

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1 year ago

Use the Trapezoidal Rule, the Midpoint Rule, and Simpson's Rule to approximate the given integral with the specified value of n.

Vera_Pavlovna [14]

Split up the integration interval into 4 subintervals:

$\left[0,\dfrac\pi8\right],\left[\dfrac\pi8,\dfrac\pi4\right],\left[\dfrac\pi4,\dfrac{3\pi}8\right],\left[\dfrac{3\pi}8,\dfrac\pi2\right]$

The left and right endpoints of the $i$ -th subinterval, respectively, are

$\ell_i=\dfrac{i-1}4\left(\dfrac\pi2-0\right)=\dfrac{(i-1)\pi}8$

$r_i=\dfrac i4\left(\dfrac\pi2-0\right)=\dfrac{i\pi}8$

for $1\le i\le4$ , and the respective midpoints are

$m_i=\dfrac{\ell_i+r_i}2=\dfrac{(2i-1)\pi}8$

Trapezoidal rule

We approximate the (signed) area under the curve over each subinterval by

$T_i=\dfrac{f(\ell_i)+f(r_i)}2(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4T_i\approx\boxed{3.038078}$

Midpoint rule

We approximate the area for each subinterval by

$M_i=f(m_i)(\ell_i-r_i)$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4M_i\approx\boxed{2.981137}$

Simpson's rule

We first interpolate the integrand over each subinterval by a quadratic polynomial $p_i(x)$ , where

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

so that

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It so happens that the integral of $p_i(x)$ reduces nicely to the form you're probably more familiar with,

$S_i=\displaystyle\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx=\frac{r_i-\ell_i}6(f(\ell_i)+4f(m_i)+f(r_i))$

Then the integral is approximately

$\displaystyle\int_0^{\pi/2}\frac3{1+\cos x}\,\mathrm dx\approx\sum_{i=1}^4S_i\approx\boxed{3.000117}$

Compare these to the actual value of the integral, 3. I've included plots of the approximations below.

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stealth61 [152]

There is an error in clever.

<h3>What is the history of Angkor?</h3>

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To learn more about Angkor, Visit:

brainly.com/question/486956

#SPJ4

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