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Would be really helpful!

Would be really helpful!-example-1
User Notacouch
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1 Answer

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Explanation:

To solve this problem, we need to use the product rule of differentiation and some trigonometric identities. Let's start by finding the derivative of y with respect to x:

y = (sin 2x) √(3+2x)

Using the product rule, we get:

dy/dx = (sin 2x) d/dx(√(3+2x)) + (√(3+2x)) d/dx(sin 2x)

To find these derivatives, we need to use the chain rule and the derivative of sin 2x:

d/dx(√(3+2x)) = (1/2√(3+2x)) d/dx(3+2x) = (1/√(3+2x))

d/dx(sin 2x) = 2cos 2x

Substituting these values, we get:

dy/dx = (sin 2x) / √(3+2x) + 2cos 2x (√(3+2x))

Now, we need to simplify this expression to the desired form. To do that, we can use the trigonometric identity:

sin 2x = 2sin x cos x

Substituting this value, we get:

dy/dx = 2sin x cos x / √(3+2x) + 2cos 2x (√(3+2x))

Now, we can use the trigonometric identity:

cos 2x = 1 - 2sin^2 x

Substituting this value, we get:

dy/dx = 2sin x cos x / √(3+2x) + 2(1 - 2sin^2 x)(√(3+2x))

Simplifying further, we get:

dy/dx = (2cos x - 4cos x sin^2 x) / √(3+2x) + 2√(3+2x) - 4sin^2 x√(3+2x)

Now, we can see that this expression matches the desired form:

dy/dx = sin 2x + (4 + Bx)cos 2x / √(3+2x)

where A = -4 and B = -2. Therefore, we have shown that:

dy/dr = sin 2x + (4 - 2x)cos 2x / √(3+2x)

where A = -4 and B = -2.

User Daniel Hawkins
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