Solve: (19 + 1^5) / 1/2 +5

Help me with differentation and integration please!!

Marina86 [1]

Answer:

See below

Step-by-step explanation:

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

Recall

$\dfrac{d}{dx}\tan x=\sec^2$

Using the chain rule

$\dfrac{dy}{dx}= \dfrac{dy}{du} \dfrac{du}{dx}$

such that $u = \tan x$

we can get a general formulation for

$y = \tan^n x$

Considering the power rule

$\boxed{\dfrac{d}{dx} x^n = nx^{n-1}}$

we have

$\dfrac{dy}{dx} =n u^{n-1} \sec^2 x \implies \dfrac{dy}{dx} =n \tan^{n-1} \sec^2 x$

therefore,

$\dfrac{d}{dx}\tan^3 x=3\tan^2x \sec^2x$

Now, once

$\sec^2 x - 1= \tan^2x$

we have

$3\tan^2x \sec^2x = 3(\sec^2 x - 1) \sec^2x = 3\sec^4x-3\sec^2x$

Hence, we showed

$\dfrac{d}{dx} (\tan^3 x) = 3\sec^4 x - 3\sec^2 x$

================================================

For the integration,

$\int \sec^4 x\, dx$

considering the previous part, we will use the identity

$\boxed{\sec^2 x - 1= \tan^2x}$

thus

$\int\sec^4x\,dx=\int \sec^2 x(\tan^2x+1)\,dx = \int \sec^2 x \tan^2x+\sec^2 x\,dx$

and

$\int \sec^2 x \tan^2x+\sec^2 x\,dx = \int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx$

Considering $u = \tan x$

and then $du=\sec^2x\ dx$

we have

$\int u^2 \, du = \dfrac{u^3}{3}+C$

Therefore,

$\int \sec^2 x \tan^2x\,dx + \int \sec^2 x\,dx = \dfrac{\tan^3 x}{3}+\tan x + C$

$\boxed{\int \sec^4 x\, dx = \dfrac{\tan^3 x}{3}+\tan x + C }$

6 0

2 years ago

Pls helppppppppppp

sattari [20]

Answer:7/4 - 7/8 = 14/8 - 7/8 = 7/8 for lettuce

Step-by-step explanation:1/4 + 5/8 = 2/8 + 5/8 = 7/8

= amount used for pepper and tomatoes

2 1/2 - 3/4 = 5/2-3/4 = 10/4 -3/4 = 7/4

= total amount used

so

7/4 - 7/8 = 14/8 - 7/8 = 7/8 for lettuce

6 0

2 years ago

Read 2 more answers

Carole's car averages 13.2 mi/gal in city driving and 25.3 mi/gal in highway driving. If she drove a total of 396.0 mi on 19 gal

nika2105 [10]

Answer:

7 gallons were used for city driving.

Step-by-step explanation:

I am going to solve this question by a system of equations.

I am going to say that:

x is the number of gallons driven in the city.

y is the number of gallons driven in the highway.

19 gal of gas

This means that $x + y = 19$

Carole's car averages 13.2 mi/gal in city driving and 25.3 mi/gal in highway driving. If she drove a total of 396.0 mi.

This means that $13.2x + 25.3y = 396$

How many of the gallons (to the nearest gallon) were used for city driving?

This is x

From the first equation, i will have

$y = 19 - x$

Replacing in the second equation

$13.2x + 25.3y = 396$

$13.2x + 25.3(19 - x) = 396$

$13.2x - 25.3x = 396 - 480,7$

$12.1x = 84.7$

$x = \frac{84.7}{12.1}$

$x = 7$

7 gallons were used for city driving.

5 0

2 years ago

Please help !!! Number 11

romanna [79]

Answer:

C) 153.9 in²

Step-by-step explanation:

14 ÷ 2 = 7

π = 3.14

3.14×7×7 = 153.9 in²

hope this helps!

5 0

2 years ago

Read 2 more answers

Determine which of the lines if any are parallel or perpendicular?

Tcecarenko [31]

Answer:

y=6x-2 is perpendicular to the 6y=-x+7

y=6x+1 is parallel to the y=6x-2

Step-by-step explanation:

if product of two lines gradient is -1 than lines are perpendicular

gr y=6x-2

y=mx+c m=gradient

1)m=6

2)m=-1/6

3)m=6

if gradients of 2 lines are equal they are paralel

4 0

2 years ago