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 February 18th, 2019, 09:20 PM #1 Newbie   Joined: Feb 2019 From: N/A Posts: 18 Thanks: 0 Math Focus: Calculus Derivative problems help Here are some of my practice problems: csc(y^2)=5x+4 and 2x^2+3=e^(2y)^3 Can you show me the steps on each one individually in detail so I can better understand how to do it? February 18th, 2019, 09:30 PM #2 Senior Member   Joined: Sep 2015 From: USA Posts: 2,530 Thanks: 1390 are you trying to find $\dfrac{dy}{dx}$ ? February 18th, 2019, 09:48 PM   #3
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Quote:
 Originally Posted by romsek are you trying to find $\dfrac{dy}{dx}$ ?
yes February 18th, 2019, 09:58 PM   #4
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Quote:
 Originally Posted by MikeAndIke Here are some of my practice problems: csc(y^2)=5x+4 and 2x^2+3=e^(2y)^3 Can you show me the steps on each one individually in detail so I can better understand how to do it?
$\csc(y^2) = 5x + 4$

$\text{Take derivatives with respect to }x \text{ on boths sides.}$

$-\cot(y^2)\csc(y^2)2y y^\prime = 5$

$y^\prime = \dfrac{5}{-2y\cot(y^2)\csc(y^2)} =$

$-\dfrac{5}{2y} \tan(y^2) \sin(y^2)$

-----------
$2x^2+3 = e^{(2y)^3} = e^{8y^3}$

$4x = e^{8y^3}24y^2y^\prime$

$y^\prime = \dfrac{x}{6e^{8y^3}y^2}$ February 18th, 2019, 10:06 PM   #5
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Quote:
 Originally Posted by romsek $\csc(y^2) = 5x + 4$ $\text{Take derivatives with respect to }x \text{ on boths sides.}$ $-\cot(y^2)\csc(y^2)2y y^\prime = 5$ $y^\prime = \dfrac{5}{-2y\cot(y^2)\csc(y^2)} =$ $-\dfrac{5}{2y} \tan(y^2) \sin(y^2)$ ----------- $2x^2+3 = e^{(2y)^3} = e^{8y^3}$ $4x = e^{8y^3}24y^2y^\prime$ $y^\prime = \dfrac{x}{6e^{8y^3}y^2}$
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