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liq [111]
3 years ago
6

Derivative of sin(x+y)

Mathematics
1 answer:
vlabodo [156]3 years ago
4 0

Answer:

\frac{dy}{dx}  =  \frac{ \cos(x + y)}{[1 - \cos(x + y)] }

Step-by-step explanation:

Let y = sin(x+y)

Differentiating with respect to x on both sides.

\frac{dy}{dx}  =  \frac{d}{dx}  \sin(x + y) \\  \\  \frac{dy}{dx}  =  \cos(x + y)\frac{d}{dx}  (x + y) \\  \\ \frac{dy}{dx}  =  \cos(x + y)(1 + \frac{dy}{dx}) \\  \\  \frac{dy}{dx}  =  \cos(x + y) +\cos(x + y) \frac{dy}{dx} \\  \\  \frac{dy}{dx}   - \cos(x + y) \frac{dy}{dx} =  \cos(x + y) \\  \\  \frac{dy}{dx}  [1 - \cos(x + y)] =  \cos(x + y) \\  \\ \purple {\bold {\frac{dy}{dx}  =  \frac{ \cos(x + y)}{[1 - \cos(x + y)] }}}

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Which expression is equivalent to (16 x Superscript 8 Baseline y Superscript negative 12 Baseline) Superscript one-half?.
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To solve the problem we must know the Basic Rules of Exponentiation.

<h2>Basic Rules of Exponentiation</h2>
  • x^ax^b = x^{(a+b)}
  • \dfrac{x^a}{x^b} = x^{(a-b)}
  • (a^a)^b =x^{(a\times b)}
  • (xy)^a = x^ay^a
  • x^{\frac{3}{4}} = \sqrt[4]{x^3}= (\sqrt[3]{x})^4

The solution of the expression is \dfrac{4x^4}{y^6}.

<h2>Explanation</h2>

Given to us

  • (16x^8y^{12})^{\frac{1}{2}}

Solution

We know that 16 can be reduced to 2^4,

=(2^4x^8y^{12})^{\frac{1}{2}}

Using identity (xy)^a = x^ay^a,

=(2^4)^{\frac{1}{2}}(x^8)^{\frac{1}{2}}(y^{12})^{\frac{1}{2}}

Using identity (a^a)^b =x^{(a\times b)},

=(2^{4\times \frac{1}{2}})\ (x^{8\times\frac{1}{2}})\ (y^{12\times{\frac{1}{2}}})

Solving further

=2^2x^4y^{-6}

Using identity \dfrac{x^a}{x^b} = x^{(a-b)},

=\dfrac{2^2x^4}{y^6}

=\dfrac{4x^4}{y^6}

Hence, the solution of the expression is \dfrac{4x^4}{y^6}.

Learn more about Exponentiation:

brainly.com/question/2193820

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2 years ago
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