Y=(2x-3)^5(2-x^4)^3 differentiate the function please

1 answer:

5 0

$\bf y=(2x-3)^5(2-x^4)^3\\\\
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\cfrac{dy}{dx}=\stackrel{\textit{product rule}}{[5(2x-3)^4\cdot 2(2-x^4)3]~~+~~[(2x-3)^5[3(2-x^4)^2(-4x^3)]]}
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\cfrac{dy}{dx}=10(2x-3)^4(2-x^4)^3~~-~~12x^3(2x-3)^5(2-x^4)^2
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\cfrac{dy}{dx}=\stackrel{\textit{common factor}}{2(2x-3)^4(2-x^4)^2}~[5(2-x^4)-6x^3(2x-3)]
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\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2~[10-5x^4-12x^4+18x^3]
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\cfrac{dy}{dx}=2(2x-3)^4(2-x^4)^2(10-17x^4+18x^3)$

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4 0

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Tickets for the basketball game were sold at $4.00 for adults and $2.50 for students. If 320 tickets were sold for a total of $1

makkiz [27]

The problem based on condition are solved using the unknown variables. The number of tickets sold to the student are 120 and the number of tickets sold to the adults are 200.

Given information-

The price of the tickets for the basketball game for adults is $4.00.

The price of the tickets for the basketball game for students is $2.50.

<h3>Variables</h3>

Variables are the unknown and the value of the variables depend on the other variables in the equation. The above problem can be solved defining the variables from the given condition.

Let the total tickets purchased by the students is<em> x</em> and the total tickets porches by the adults is <em>y.</em>

As the total tickets sold for the game is $320. Thus,

$\begin{aligned}\\
x+y&=320\\
y&=320-x\\
\end$ .......1