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Anvisha [2.4K]
2 years ago
7

C%3A%20%20is%20%20%5C%3A%20doubled%20%20%5C%3A%20then%20%20%5C%3A%20%20%E2%80%8B%7D" id="TexFormula1" title=" \sf{prove \: that \: if \: a \: number \: is \: doubled \: then \: ​}" alt=" \sf{prove \: that \: if \: a \: number \: is \: doubled \: then \: ​}" align="absmiddle" class="latex-formula">
\sf{its  \: cube \:  is \:  8 \:  times  \: the \:  cube \:  of \:  the  \: given  \: number}
__________________​
Mathematics
2 answers:
IgorLugansk [536]2 years ago
7 0

\underline{\Large\underline\frak {\underline\red{Given:-} }}

A number is doubled then it's Cube is 8 times.

\underline{\Large\underline\frak {\underline\orange{Vube \:figure } } :- }

\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered} {\begin{gathered} \sf{ } \: \: \: \huge\boxed{ \begin{array}{cc} \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \end{array}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf{ \: } \end{gathered}}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}

\begin{gathered}\underline{\Large\underline\frak {\underline\purple{ Formula \:used \: \: } : }} \\ \\ \sf \: \blue{\underline{\boxed{ \sf \: a \times a \times a = {a}^{3} }}}\end{gathered}

\bf \: Now\bf \: according \: to \: Question :-

\begin{gathered} \sf \: double \: of \: cube \: = 2 + a \\ \\ \sf \qquad \qquad \: \: \: \: \: \: \: = 2a\end{gathered}

\begin{gathered} \sf \: let \: a \: number \: be \: = a \\ \\ \sf \: it's \: cube \: will \: be \: = a \times a \times a = {a}^{3} \\ \end{gathered}

\sf \pink{  = 2a \times 2a \times 2a = 8 {a}^{2} \: or \: {(2a)}^{3}}

Hence, Proved that double 8 times the cube of a number = 8 times the cube of number a = 8 × a³ = 8 × x³

\bf \green{R.H.S. = L.H.S.}

<h2>______________</h2>
lora16 [44]2 years ago
6 0

<h3>let a number = a</h3><h3>it's cube will be a×a×a = a³</h3>

<h3>now atq 2a </h3><h3>it's cube will be </h3><h3>2a ×2a ×2a => 8a³ or (2a)³</h3>

<h2>Hope it helps you </h2>
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If you're using the app, try seeing this answer through your browser:  brainly.com/question/3000160

————————

Find the derivative of

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You can treat  y  as a composite function of  x:

\left\{\! \begin{array}{l} \mathsf{y=\ell n\,u}\\\\ \mathsf{u=\dfrac{1+sin\,x}{cos\,x}} \end{array} \right.


so use the chain rule to differentiate  y:

\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(\ell n\,u)\cdot \dfrac{d}{dx}\!\left(\dfrac{1+sin\,x}{cos\,x}\right)}


The first derivative is  1/u, and the second one can be evaluated by applying the quotient rule:

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{\frac{d}{dx}(1+sin\,x)\cdot cos\,x-(1+sin\,x)\cdot \frac{d}{dx}(cos\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(0+cos\,x)\cdot cos\,x-(1+sin\,x)\cdot (-\,sin\,x)}{(cos\,x)^2}}


Multiply out those terms in parentheses:

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


Substitute back for  \mathsf{u=\dfrac{1+sin\,x}{cos\,x}:}

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{~\frac{1+sin\,x}{cos\,x}~}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{cos\,x}{1+sin\,x}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}


Simplifying that product, you get

\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+sin\,x}\cdot \dfrac{1+sin\,x}{cos\,x}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{cos\,x}}


∴     \boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=sec\,x} \end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}


I hope this helps. =)


Tags:  <em>derivative composite function logarithmic logarithm log trigonometric trig secant tangent sec tan chain rule quotient rule differential integral calculus</em>

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Step 2: Divide both sides by -16.

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