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3 years ago

Explain what chain rule is . workout appropriate examples to support your explanation.

1 answer:

5 0

The chain rule is a formula for computing the derivative of the composition of two or more functions.

 (f o g(x))1=g′(x)f′(g(x))
or
 dy/dx=dy/du du/dx

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What is the magnitude of vector K ?

vovangra [49]

Answer:

Vector K = 8 m

Explanation:

The given figure shows a right angle triangle JKL. It is given that :

vector L = 10 m

vector J = 6 m

We have to find vector k. We can find it by using Pythagoras theorem. According to this theorem, the sum of squares of perpendicular and the base is equal to the square of the longest side.

$L^2=K^2+J^2$

$K^2=L^2-J^2$

$K^2=(100)^2-(6)^2$

K = 8 m

Hence, the magnitude of vector K is 8 m.

3 0

3 years ago

A student wants to determine whether an unknown solid is an ionic compound. He puts some in water and finds that it does not dis

elena-14-01-66 [18.8K]

Answer:

Explanation:

8 0

3 years ago

A football is thrown with an acceleration of 15 m/s^2 and a force of 13 N. What is its mass?

-Dominant- [34]

I believe it is b. Lmk if I’m wrong

7 0

3 years ago

Read 2 more answers

A speaker is designed for wide dispersion for a high frequency sound. What should the diameter of the circular opening be in a s

Greeley [361]

Answer:

$a = 4.76 cm$

Explanation:

As we know by the law of diffraction through circular aperture

$a sin\theta = 1.22 \lambda$

here we know that

a = diameter of the aperture

$\theta$ = diffraction angle

$\lambda$ = wavelength

now we have

$a = \frac{1.22 \lambda}{sin\theta}$

Now in order to find the wavelength we know

$\lambda = \frac{c}{f}$

$\lambda = \frac{343}{9100}$

$\lambda = 0.0376 m$

now we have

$a = \frac{1.22(0.0376)}{sin75}$

$a = \frac{0.046}{0.966}$

$a = 0.0476 m= 4.76 cm$

6 0

3 years ago

How much work did the movers do (horizontally) pushing a 41.0-kg crate 10.6 m across a rough floor without acceleration, if the

VLD [36.1K]

Answer:

The required work done is $2555.448~J$

Explanation:

Consider 'F' is the applied force on the crate and 'f' be the force created by friction. According to the figure if ' $\mu_{k}$ ' be the coefficient of friction, then

$f = \mu_{k} \times N = \mu_{k} \times Mg$

where 'M', 'N' and 'g' are the mass of the crate, the normal force aced upon the block and the acceleration due to gravity respectively.

Since the application of force by the movers does not create any acceleration to the block, we can write

$F = f = \mu_{k} \times M \times g = 0.6 \times 41~Kg~ \times 9.8~m~s^{-2} = 241.08~N$

So the work done (W) in moving the crate by a distance s = 10.6 m is

$W = F \times s = 241.08~N \times 10.6~m = 2555.448 J$

5 0

3 years ago