All categories

2 years ago

Differentiate y=xln(3x)

Mathematics

2 answers:

OLEGan [10]2 years ago

7 0

Can you explain more ?

Ganezh [65]2 years ago

6 0

Answer:

Remove parentheses.

y=xln3x

2 Already solved. No work needed.

y=xln3x

the solution is in the picture above please mark me brainliest:)

You might be interested in

Write an equation of the parabola with the vertex (-5,4) and focus (-5,0)

ch4aika [34]

Answer:

-0.4 or -10.4

Step-by-step explanation:

5 0

2 years ago

What is the value of i^3?

Pavel [41]

Answer:

the answer is -i

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

It’s a easy mathswatch questionnnnn helpppppp meeeeeeeee

pychu [463]

Answer:

15, 6 and - 3

Step-by-step explanation:

Given x - y = 3 , then

(a)

5(x - y) = 5(3) = 15

(b)

2x - 2y ← factor out 2 from each term

= 2(x - y) = 2(3) = 6

(c)

y - x ← factor out - 1 from each term

= - 1(x - y) = - 1(3) = - 3

8 0

2 years ago

Ana is planning to plant roses around the edge of a circular flowerbed having a diameter of 7 feet. If the total cost of plantin

m_a_m_a [10]

I forget how to do this. I haven't done this in awhile.

8 0

3 years ago

Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustra

Rainbow [258]

Answer:

$x = t$

$y = 1 - t$

$z = 2t$

Step-by-step explanation:

Given

$x=t$

$y=e^{-t}$

$z=2t-t^2$

(0, 1, 0)

The vector equation is given as:

$r(t) = (x,y,z)$

Substitute values for x, y and z

$r(t) = (t,\ e^{-t},\ 2t - t^2)$

Differentiate:

$r'(t) = (1,\ -e^{-t},\ 2 - 2t)$

The parametric value that corresponds to (0, 1, 0) is:

$t = 0$

Substitute 0 for t in r'(t)

$r'(t) = (1,\ -e^{-t},\ 2 - 2t)$

$r'(0) = (1,\ -e^{-0},\ 2 - 2*0)$

$r'(0) = (1,\ -1,\ 2 - 0)$

$r'(0) = (1,\ -1,\ 2)$

The tangent line passes through (0, 1, 0) and the tangent line is parallel to r'(0)

It should be noted that:

The equation of a line through position vector a and parallel to vector v is given as:

$r(t) = a + tv$

Such that:

$a = (0,1,0)$ and $v = r'(0) = (1,-1,2)$

The equation becomes:

$r(t) = (0,1,0) + t(1,-1,2)$

$r(t) = (0,1,0) + (t,-t,2t)$

$r(t) = (0+t,1-t,0+2t)$

$r(t) = (t,1-t,2t)$

By comparison:

$r(t) = (x,y,z)$ and $r(t) = (t,1-t,2t)$

The parametric equations for the tangent line are:

$x = t$

$y = 1 - t$

$z = 2t$

7 0

2 years ago

Differentiate y=xln(3x)​

Differentiate y=xln(3x)