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(8 pts) Using implicit differentiation, find
dy//dx for the curve

2cos(4x^(3))sin(7y)=-5x

$1$ . ( $8$ pts) Using implicit differentiation, find $d y / d x$ for the curve $\newline$ $2 \cos \left(4 x^{3}\right) \sin (7 y)=-5 x$

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Find derivatives using logarithmic differentiation

Full solution

Q.

1

. (

8

pts) Using implicit differentiation, find

d y / d x

for the curve

\newline

2 \cos \left(4 x^{3}\right) \sin (7 y)=-5 x

Differentiate Left Side: To find $\frac{dy}{dx}$ using implicit differentiation, we need to differentiate both sides of the equation with respect to $x$ , treating $y$ as a function of $x$ ( $y=y(x)$ ). The equation is $2\cos(4x^3)\sin(7y) = -5x$ .
Apply Product Rule: Differentiate the left side of the equation with respect to $x$ . We will use the product rule, which states that the derivative of a product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. We also need to use the chain rule for each function.
Differentiate Right Side: Differentiating $2\cos(4x^3)$ with respect to $x$ gives us $-2 \times 3 \times 4x^2 \times \sin(4x^3)$ because the derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$ , and we then multiply by the derivative of the inside function $4x^3$ , which is $12x^2$ .
Equate Derivatives: Differentiating $\sin(7y)$ with respect to $x$ gives us $7\cos(7y) \cdot \frac{dy}{dx}$ because the derivative of $\sin(u)$ with respect to $u$ is $\cos(u)$ , and we then multiply by the derivative of the inside function $7y$ with respect to $x$ , which is $7 \cdot \frac{dy}{dx}$ .
Isolate Term for $\frac{dy}{dx}$ : Now we apply the product rule: the derivative of $2\cos(4x^3)\sin(7y)$ with respect to $x$ is $(-24x^2 \cdot \sin(4x^3)) \cdot \sin(7y) + 2\cos(4x^3) \cdot (7\cos(7y) \cdot \frac{dy}{dx})$ .
Simplify Equation: Differentiate the right side of the equation with respect to $x$ , which is $-5x$ . The derivative of $-5x$ with respect to $x$ is $-5$ .
Divide by Constants: Now we equate the derivatives from both sides of the equation: $(-24x^2 \cdot \sin(4x^3)) \cdot \sin(7y) + 2\cos(4x^3) \cdot (7\cos(7y) \cdot \frac{dy}{dx}) = -5$ .
Divide by Constants: Now we equate the derivatives from both sides of the equation: $(-24x^2 \cdot \sin(4x^3)) \cdot \sin(7y) + 2\cos(4x^3) \cdot (7\cos(7y) \cdot \frac{dy}{dx}) = -5$ .We need to solve for $\frac{dy}{dx}$ . To do this, we isolate the term containing $\frac{dy}{dx}$ on one side of the equation. This gives us $2\cos(4x^3) \cdot (7\cos(7y) \cdot \frac{dy}{dx}) = 5 - (-24x^2 \cdot \sin(4x^3)) \cdot \sin(7y)$ .
Divide by Constants: Now we equate the derivatives from both sides of the equation: $(-24x^2 \cdot \sin(4x^3)) \cdot \sin(7y) + 2\cos(4x^3) \cdot (7\cos(7y) \cdot \frac{dy}{dx}) = -5$ .We need to solve for $\frac{dy}{dx}$ . To do this, we isolate the term containing $\frac{dy}{dx}$ on one side of the equation. This gives us $2\cos(4x^3) \cdot (7\cos(7y) \cdot \frac{dy}{dx}) = 5 - (-24x^2 \cdot \sin(4x^3)) \cdot \sin(7y)$ .Simplify the equation to solve for $\frac{dy}{dx}$ : $14\cos(4x^3)\cos(7y) \cdot \frac{dy}{dx} = 5 + 24x^2 \cdot \sin(4x^3) \cdot \sin(7y)$ .
Divide by Constants: Now we equate the derivatives from both sides of the equation: $(-24x^2 \cdot \sin(4x^3)) \cdot \sin(7y) + 2\cos(4x^3) \cdot (7\cos(7y) \cdot \frac{dy}{dx}) = -5$ .We need to solve for $\frac{dy}{dx}$ . To do this, we isolate the term containing $\frac{dy}{dx}$ on one side of the equation. This gives us $2\cos(4x^3) \cdot (7\cos(7y) \cdot \frac{dy}{dx}) = 5 - (-24x^2 \cdot \sin(4x^3)) \cdot \sin(7y)$ .Simplify the equation to solve for $\frac{dy}{dx}$ : $14\cos(4x^3)\cos(7y) \cdot \frac{dy}{dx} = 5 + 24x^2 \cdot \sin(4x^3) \cdot \sin(7y)$ .Finally, divide both sides by $14\cos(4x^3)\cos(7y)$ to get $\frac{dy}{dx}$ on its own: $\frac{dy}{dx} = \frac{5 + 24x^2 \cdot \sin(4x^3) \cdot \sin(7y)}{14\cos(4x^3)\cos(7y)}$ .

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