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andre [41]
3 years ago
6

Find Derivative (root3(6x^2))

Mathematics
2 answers:
mars1129 [50]3 years ago
7 0
This is the same as saying (6x^2)^3. We can take the derivative using the chain rule. With the chain rule, you multiply by the power, decrease the power by 1, and multiply by the derivative of the inside.

3(6x^2)^2*12x
or
36x*(6x^2)^2

Hope this helps<span />
trapecia [35]3 years ago
5 0
\bf \sqrt[3]{6x^2}\implies (6x^2)^{\frac{1}{3}}\implies 6^{\frac{1}{3}}x^{\frac{2}{3}}\\\\&#10;-----------------------------\\\\&#10;\cfrac{dy}{dx}=6^{\frac{1}{3}}\cdot \cfrac{2}{3}x^{\frac{2}{3}-1}\implies 6^{\frac{1}{3}}\cdot \cfrac{2}{3}x^{-\frac{1}{3}}\implies 6^{\frac{1}{3}}\cdot \cfrac{2}{3}\cdot \cfrac{1}{x^{\frac{1}{3}}}&#10;\\\\\\&#10;

\bf \cfrac{dy}{dx}=6^{\frac{1}{3}}\cdot \cfrac{2}{3x^{\frac{1}{3}}}\implies \cfrac{2\cdot 6^{\frac{1}{3}}}{3x^{\frac{1}{3}}}\implies \cfrac{2\sqrt[3]{6}}{3\sqrt[3]{x}}\implies \cfrac{2}{3}\sqrt[3]{\cfrac{6}{x}}
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Can you guys help me please as much as you can ^^
GalinKa [24]

Answer:

(m1) =  50

(m2)=

(m3)= 108

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(m5)=

(m6)=

(m7)=

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(m10)= 18

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3 0
2 years ago
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Using factor theorem , show that (x -4) is a factor 2x3 + x2 -26x - 40 and hence factorize completely.
notka56 [123]

Step-by-step explanation:

if x-4 is a factor of 2x^3 + x^2- 26x - 40

then f(4) = 0

f(x) = 2x^3 + x^2- 26x - 40

f(4) = 2(4)^3 + (4)^2 - 26(4) - 40

f(4)= 2(64) + 16 - 104 - 40

f(4) = 128 + 16 - 104 - 40

f(4) = 0

hence factorize completely is the photo

7 0
3 years ago
Consider the curve defined by the equation y=6x2+14x. Set up an integral that represents the length of curve from the point (−2,
torisob [31]

Answer:

32.66 units

Step-by-step explanation:

We are given that

y=6x^2+14x

Point A=(-2,-4) and point B=(1,20)

Differentiate w.r. t x

\frac{dy}{dx}=12x+14

We know that length of curve

s=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}dx

We have a=-2 and b=1

Using the formula

Length of curve=s=\int_{-2}^{1}\sqrt{1+(12x+14)^2}dx

Using substitution method

Substitute t=12x+14

Differentiate w.r t. x

dt=12dx

dx=\frac{1}{12}dt

Length of curve=s=\frac{1}{12}\int_{-2}^{1}\sqrt{1+t^2}dt

We know that

\sqrt{x^2+a^2}dx=\frac{x\sqrt {x^2+a^2}}{2}+\frac{1}{2}\ln(x+\sqrt {x^2+a^2})+C

By using the formula

Length of curve=s=\frac{1}{12}[\frac{t}{2}\sqrt{1+t^2}+\frac{1}{2}ln(t+\sqrt{1+t^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}[\frac{12x+14}{2}\sqrt{1+(12x+14)^2}+\frac{1}{2}ln(12x+14+\sqrt{1+(12x+14)^2})]^{1}_{-2}

Length of curve=s=\frac{1}{12}(\frac{(12+14)\sqrt{1+(26)^2}}{2}+\frac{1}{2}ln(26+\sqrt{1+(26)^2})-\frac{12(-2)+14}{2}\sqrt{1+(-10)^2}-\frac{1}{2}ln(-10+\sqrt{1+(-10)^2})

Length of curve=s=\frac{1}{12}(13\sqrt{677}+\frac{1}{2}ln(26+\sqrt{677})+5\sqrt{101}-\frac{1}{2}ln(-10+\sqrt{101})

Length of curve=s=32.66

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3 years ago
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Answer: 28 in.

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