Find the derivative of f(x) = (1 + 5x2)(x x2) in two ways. use the product

Find the derivative of f(x) = (1 + 5x2)(x − x2) in two ways. use the product rule. f '(x) = perform the multiplication first. f '(x) = do your answers agree? yes no

2 months ago

Solution 1

Guest Guest #10154
2 months ago

Answer:

The answers agree.

f^{\prime}(x) = - 20x^{3} + 15x^{2} - 2x + 1

Step-by-step explanation:

The product rule is:

f(x) = g(x)h(x)

f^{\prime}(x) = g^{\prime}(x)h(x) + g(x)h^{\prime}(x)

In this problem, we have that:

f(x) = (1 + 5x^{2})(x - x^{2})

So

g(x) = 1 + 5x^{2}

g^{\prime}(x) = 10x

h(x) = x - x^{2}

h^{\prime}(x) = 1 - 2x.

f^{\prime}(x) = g^{\prime}(x)h(x) + g(x)h^{\prime}(x)

f^{\prime}(x) = (10x)*(x - x^{2}) + (1 + 5x^{2})(1-2x)

f^{\prime}(x) = 10x^{2} - 10x^{3} + 1 - 2x + 5x^{2} - 10x^{3}

f^{\prime}(x) = - 20x^{3} + 15x^{2} - 2x + 1

Multiplication first:

f(x) = (1 + 5x^{2})(x - x^{2})

f(x) = x - x^{2} + 5x^{3} - 5x^{4}

f(x) = -5x^{4} + 5x^{3} - x^{2} + x

f^{\prime}(x) = -20x^{3} + 15x^{2} - 2x + 1

Solution 2

Guest Guest #10155
2 months ago
The product rule states that given a function:
f(x)=g(x)h(x)
f'(x)=g'(x)h(x)+h'(x)g(x)
thus the derivative of the expression will be given by:
 f(x) = (1 + 5x2)(x − x2)
g(x)=(1+5x^2)
g'(x)=10x

h(x)=(x-x^2)
h'(x)=1-2x
thus:
f'(x)=10x(x-x^2)+(1-2x)(1+5x^2)
simplifying this we get:
f'(x)=-20x^3+15x^2-12x+1

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Step-by-step explanation:

Let r(u,v)=\left \langle u\cos v,u\sin v,u \right \rangle

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r_u\times r_v=\left | \begin{matrix} i&j&k\\ \cos v&\sin v&1\\-u\sin v&u\cos v&0 \end{matrix} \right |=\left \langle -u\cos v,-u\sin v,u \right \rangle\\\left \|r_u\times r_v  \right \|=\sqrt{u^2\cos ^2v+u^2\sin ^2v+u^2}\\=\sqrt{u^2\left (\cos ^2v+\sin ^2v  \right )+u^2}\\=\sqrt{u^2+u^2}\\=\sqrt{2u^2}=\sqrt{2}u

dS=\left \| r_u\times r_v \right \|\,du\,\,dv

\int \int f\,dS=\int_v \int_u f\,\left \| r_u\times r_v \right \|\,du\,\,dv

To find f, put x = u \cos(v)\,,\, y = u \sin(v)\,,\, z = u in xyz.

f=xyz\\=(u \cos(v))( u \sin(v))(u)\\=u^3\sin v\,\cos v\\=u^3\left ( \frac{2\sin v\,\cos v}{2} \right )\\=\frac{u^3}{2}\sin 2v\,\,\left \{ \because 2\sin v\,\cos v=\sin 2v \right \}

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\int \int f\,dS=\int_{0}^{\frac{\pi}{2}} \int_{0}^{3} \frac{u^3}{2}\sin 2v\,\left ( \sqrt{2}u \right )\,du\,\,dv\\=\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}} \int_{0}^{3}u^4\sin 2v\,\,du\,\,dv

Integrate with respect to u.

\int \int f\,dS=\frac{\sqrt{2}}{2}\int_{0}^{\frac{\pi}{2}}\sin 2v\left [ \frac{u^5}{5} \right ]_{0}^{3}\,\,dv\,\,\left \{ \because \int u^n\,du=\frac{u^{n+1}}{n+1} \right \}\\=\frac{\sqrt{2}}{10}(243)\int_{0}^{\frac{\pi}{2}}\sin 2v\,\,dv

Integrate with respect to v.

\int \int f\,dS=\frac{243\sqrt{2}}{10}\left [ \frac{-\cos 2v}{2} \right ]_{0}^{\frac{\pi}{2}}\,\,\left \{ \because \int \sin v=-\cos v \right \}\\=\frac{243\sqrt{2}}{20}\left ( -1-1 \right )\\=\frac{243\sqrt{2}}{10}

Solution 2
Let's capture the parameterization of the surface \mathcal S by the vector function

\mathbf s(u,v)=\langle u\cos v,u\sin v,u\rangle

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