Ask question

Ask question

All categories

3 years ago

9

Find dy, dx if f(x) = (x + 8)3x. 3xln(x + 8) the product of the quantity 3 times the natural log of the quantity x plus 8 plus 3

times x divided by the quantity x plus 8, and the quantity x plus 8 raised to the 3x power 3 times the natural log of the quantity x plus 8 plus 3 times x divided by the quantity x plus 8 3x(x + 8)(3x − 1)

2 answers:

boyakko [2]3 years ago

8 0

F(x) = 9x^2(x + 8)^2

guajiro [1.7K]3 years ago

6 0

Answer:

$\frac{dy}{dx}=\frac{3}{x+8}$

Step-by-step explanation:

Given : The function is $f(x)=3 \ln(x+8)$

To find : The value of $\frac{dy}{dx}$ ?

Solution :

Let the function be $y=3 \ln(x+8)$

Derivate w.r.t 'x',

$\frac{dy}{dx}=\frac{d}{dx}(3 \ln(x+8))$

$\frac{dy}{dx}=3\times\frac{1}{x+8}\times 1$

$\frac{dy}{dx}=\frac{3}{x+8}$

Therefore, the value is $\frac{dy}{dx}=\frac{3}{x+8}$ .

You might be interested in

What is the size of angle x

mixer [17]

Step-by-step explanation:

180-120 = 60 which is the 2nd angle in triangle.

60+50+x = 180

x = 70

5 0

2 years ago

(-9m+9) + (8m+8) simplify

lisabon 2012 [21]

Answer:

M+16m

Step-by-step explanation:

-9m+9 cancel eachother out because it is a negative and a postive value, so you would just put the variable. For 8m+8 it would be 16m because the sre both positive so you would add them like normal

7 0

2 years ago

Read 2 more answers

A bin contains 25 light bulbs, 5 of which are in good condition and will function for at least 30 days, 10 of which are partiall

Ira Lisetskai [31]

Answer:

The probability that it will still be working after one week is $\frac{1}{5}$

Step-by-step explanation:

Given :

Total number of bulbs = 25

Number of bulbs which are good condition and will function for at least 30 days = 5

Number of bulbs which are partially defective and will fail in their second day of use = 10

Number of bulbs which are totally defective and will not light up = 10

To find : What is the probability that it will still be working after one week?

Solution :

First condition is a randomly chosen bulb initially lights,

i.e. Either it is in good condition and partially defective.

Second condition is it will still be working after one week,

i.e. Bulbs which are good condition and will function for at least 30 days

So, favorable outcome is 5

The probability that it will still be working after one week is given by,

$\text{Probability}=\frac{\text{Favorable outcome}}{\text{Total number of outcome}}$

$\text{Probability}=\frac{5}{25}$

$\text{Probability}=\frac{1}{5}$

5 0

3 years ago

What is tan 11 pie/6

tia_tia [17]

Answer:

5.75

Step-by-step explanation:

5 0

3 years ago

For each of the following vector fields F , decide whether it is conservative or not by computing curl F . Type in a potential f

Phantasy [73]

The key idea is that, if a vector field is conservative, then it has curl 0. Equivalently, if the curl is not 0, then the field is not conservative. But if we find that the curl is 0, that on its own doesn't mean the field is conservative.

1.

$\mathrm{curl}\vec F=\dfrac{\partial(5x+10y)}{\partial x}-\dfrac{\partial(-6x+5y)}{\partial y}=5-5=0$

We want to find $f$ such that $\nabla f=\vec F$ . This means

$\dfrac{\partial f}{\partial x}=-6x+5y\implies f(x,y)=-3x^2+5xy+g(y)$

$\dfrac{\partial f}{\partial y}=5x+10y=5x+\dfrac{\mathrm dg}{\mathrm dy}\implies\dfrac{\mathrm dg}{\mathrm dy}=10y\implies g(y)=5y^2+C$

$\implies\boxed{f(x,y)=-3x^2+5xy+5y^2+C}$

so $\vec F$ is conservative.

2.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(-2y)}{\partial z}-\dfrac{\partial(1)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x)}{\partial z}-\dfrac{\partial(1)}{\partial z}\right)\vec\jmath+\left(\dfrac{\partial(-2y)}{\partial x}-\dfrac{\partial(-3x)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x\implies f(x,y,z)=-\dfrac32x^2+g(y,z)$

$\dfrac{\partial f}{\partial y}=-2y=\dfrac{\partial g}{\partial y}\implies g(y,z)=-y^2+h(y)$

$\dfrac{\partial f}{\partial z}=1=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=z+C$

$\implies\boxed{f(x,y,z)=-\dfrac32x^2-y^2+z+C}$

so $\vec F$ is conservative.

3.

$\mathrm{curl}\vec F=\dfrac{\partial(10y-3x\cos y)}{\partial x}-\dfrac{\partial(-\sin y)}{\partial y}=-3\cos y+\cos y=-2\cos y\neq0$

so $\vec F$ is not conservative.

4.

$\mathrm{curl}\vec F=\left(\dfrac{\partial(5y^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial y}\right)\vec\imath+\left(\dfrac{\partial(-3x^2)}{\partial z}-\dfrac{\partial(5z^2)}{\partial x}\right)\vec\jmath+\left(\dfrac{\partial(5y^2)}{\partial x}-\dfrac{\partial(-3x^2)}{\partial y}\right)\vec k=\vec0$

Then

$\dfrac{\partial f}{\partial x}=-3x^2\implies f(x,y,z)=-x^3+g(y,z)$

$\dfrac{\partial f}{\partial y}=5y^2=\dfrac{\partial g}{\partial y}\implies g(y,z)=\dfrac53y^3+h(z)$

$\dfrac{\partial f}{\partial z}=5z^2=\dfrac{\mathrm dh}{\mathrm dz}\implies h(z)=\dfrac53z^3+C$

$\implies\boxed{f(x,y,z)=-x^3+\dfrac53y^3+\dfrac53z^3+C}$

so $\vec F$ is conservative.

4 0

3 years ago