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3 years ago

Find the derivative of y with respect to the appropriate variable y=cos^-1(x^2)

Mathematics

1 answer:

Ksju [112]3 years ago

3 0

Answer:

a. Practically speaking, you compute the differential in much the same way you compute a derivative via implicit differentiation, but you omit the variable with respect to which you are differentiating.

Aside: Compare this to what happens when you differentiate both sides with respect to some other independent parameter, say :

b. This is just a matter of plugging

Step-by-step explanation:

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ikadub [295]

Answer:

may I see a photo?

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

HELP ME PLEASE! FIRST PERSON TO ANSWER GET BRAINLIST! (where the hell its called)

ziro4ka [17]

Answer:

= 44j+32

j = -8/11

geometric figure it is called a line.

Step-by-step explanation:

Expand the following:

4 (6 j + 5 j + 8)

Grouping like terms, 5 j + 8 + 6 j = (5 j + 6 j) + 8:

4 (5 j + 6 j) + 8

5 j + 6 j = 11 j:

4 (11 j + 8)

4 (11 j + 8) = 4×8 + 4×11 j:

4×8 + 4×11 j

4×8 = 32:

32 + 4×11 j

4×11 = 44:

Answer: 32 + 44 j

7 0

3 years ago

Could someone please help me:) I am stick and I am not sure what to do

Delicious77 [7]

Answer:

Part 5.1.1:

$\displaystyle \cos 2A = \frac{7}{8}$

Part 5.1.2:

$\displaystyle \cos A = \frac{\sqrt{15}}{4}$

Step-by-step explanation:

We are given that:

$\displaystyle \sin 2A = \frac{\sqrt{15}}{8}$

Part 5.1.1

Recall that:

$\displaystyle \sin^2 \theta + \cos^2 \theta = 1$

Let θ = 2A. Hence:

$\displaystyle \sin ^2 2A + \cos ^2 2A = 1$

Square the original equation:

$\displaystyle \sin^2 2A = \frac{15}{64}$

Hence:

$\displaystyle \left(\frac{15}{64}\right) + \cos ^2 2A = 1$

Subtract:

$\displaystyle \cos ^2 2A = \frac{49}{64}$

Take the square root of both sides:

$\displaystyle \cos 2A = \pm\sqrt{\frac{49}{64}}$

Since 0° ≤ 2A ≤ 90°, cos(2A) must be positive. Hence:

$\displaystyle \cos 2A = \frac{7}{8}$

Part 5.1.2

Recall that:

$\displaystyle \begin{aligned} \cos 2\theta &= \cos^2 \theta - \sin^2 \theta \\ &= 1- 2\sin^2\theta \\ &= 2\cos^2\theta - 1\end{aligned}$

We can use the third form. Substitute:

$\displaystyle \left(\frac{7}{8}\right) = 2\cos^2 A - 1$

Solve for cosine:

$\displaystyle \begin{aligned} \frac{15}{8} &= 2\cos^2 A\\ \\ \cos^2 A &= \frac{15}{16} \\ \\ \cos A& = \pm\sqrt{\frac{15}{16}} \\ \\ \Rightarrow \cos A &= \frac{\sqrt{15}}{4}\end{aligned}$