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3 years ago

Solve.

Mathematics

2 answers:

Serhud [2]3 years ago

6 0

Answer:

Step-by-step explanation:

12.9 + 10.2x = 35.1

10.2x = 35.1 - 12.9

10.2x = 22.2

x = 22.2/10.2

x= 222/102 = 111/51

$x= 2\dfrac{9}{51}$

Vikentia [17]3 years ago

5 0

Answer:

2 9/10

Step-by-step explanation:

took test

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How to find the derivative of cos^2x? i seem to be confused.

slamgirl [31]

If you're using the app, try seeing this answer through your browser: brainly.com/question/2927231

————————

You can actually use either the product rule or the chain rule for this one. Observe:

• Method I:

y = cos² x

y = cos x · cos x

Differentiate it by applying the product rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x\cdot cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{dx}(cos\,x)\cdot cos\,x+cos\,x\cdot \dfrac{d}{dx}(cos\,x)}$

The derivative of cos x is – sin x. So you have

$\mathsf{\dfrac{dy}{dx}=(-sin\,x)\cdot cos\,x+cos\,x\cdot (-sin\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=-sin\,x\cdot cos\,x-cos\,x\cdot sin\,x}$

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

—————

• Method II:

You can also treat y as a composite function:

$\left\{\!
\begin{array}{l}
\mathsf{y=u^2}\\\\
\mathsf{u=cos\,x}
\end{array}
\right.$

and then, differentiate y by applying the chain rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\
\mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(u^2)\cdot \dfrac{d}{dx}(cos\,x)}$

For that first derivative with respect to u, just use the power rule, then you have

$\mathsf{\dfrac{dy}{dx}=2u^{2-1}\cdot \dfrac{d}{dx}(cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=2u\cdot (-sin\,x)\qquad\quad (but~~u=cos\,x)}\\\\\\
\mathsf{\dfrac{dy}{dx}=2\,cos\,x\cdot (-sin\,x)}$

and then you get the same answer:

$\therefore~~\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=-2\,sin\,x\cdot cos\,x}\end{array}}\qquad\quad\checkmark$

I hope this helps. =)

Tags: <em>derivative chain rule product rule composite function trigonometric trig squared cosine cos differential integral calculus</em>

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