Find the exact coordinates of the turning points on the curve $y=\sin^3 x\cos x$ for $0$

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Q. Find the exact coordinates of the turning points on the curve

y=\sin^3 x\cos x

for

0

Find Derivative and Critical Points: To find the turning points, we need to find the derivative of $y$ with respect to $x$ and set it equal to zero to find the critical points.
Apply Product and Chain Rule: The derivative of $y = \sin^3(x)\cos(x)$ can be found using the product rule and the chain rule. The product rule 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.
Solve for Critical Points: Let's denote $u = \sin^3(x)$ and $v = \cos(x)$ . Then, using the product rule, we have: $\newline$ $\frac{dy}{dx} = \frac{du}{dx} \cdot v + u \cdot \frac{dv}{dx}$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\newline$ $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\newline$ $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$ Solving $\frac{dv}{dx}$ $4$ gives us $\frac{dv}{dx}$ $7$ within the interval $\frac{dv}{dx}$ $8$ .
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\newline$ $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$ Solving $\frac{dv}{dx}$ $4$ gives us $\frac{dv}{dx}$ $7$ within the interval $\frac{dv}{dx}$ $8$ .Solving $\frac{dv}{dx}$ $5$ , we can use the Pythagorean identity $u = \sin^3(x)$ $0$ to rewrite the equation as: $\newline$ $u = \sin^3(x)$ $1$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\newline$ $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$ Solving $\frac{dv}{dx}$ $4$ gives us $\frac{dv}{dx}$ $7$ within the interval $\frac{dv}{dx}$ $8$ .Solving $\frac{dv}{dx}$ $5$ , we can use the Pythagorean identity $u = \sin^3(x)$ $0$ to rewrite the equation as: $\newline$ $u = \sin^3(x)$ $1$ Simplifying the equation, we get: $\newline$ $u = \sin^3(x)$ $2$ $\newline$ $u = \sin^3(x)$ $3$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\newline$ $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$ Solving $\frac{dv}{dx}$ $4$ gives us $\frac{dv}{dx}$ $7$ within the interval $\frac{dv}{dx}$ $8$ .Solving $\frac{dv}{dx}$ $5$ , we can use the Pythagorean identity $u = \sin^3(x)$ $0$ to rewrite the equation as: $\newline$ $u = \sin^3(x)$ $1$ Simplifying the equation, we get: $\newline$ $u = \sin^3(x)$ $2$ $\newline$ $u = \sin^3(x)$ $3$ Taking the square root of both sides, we get: $\newline$ $u = \sin^3(x)$ $4$
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\newline$ $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$ Solving $\frac{dv}{dx}$ $4$ gives us $\frac{dv}{dx}$ $7$ within the interval $\frac{dv}{dx}$ $8$ .Solving $\frac{dv}{dx}$ $5$ , we can use the Pythagorean identity $u = \sin^3(x)$ $0$ to rewrite the equation as: $\newline$ $u = \sin^3(x)$ $1$ Simplifying the equation, we get: $\newline$ $u = \sin^3(x)$ $2$ $\newline$ $u = \sin^3(x)$ $3$ Taking the square root of both sides, we get: $\newline$ $u = \sin^3(x)$ $4$ The values of $x$ that satisfy $u = \sin^3(x)$ $4$ within the interval $\frac{dv}{dx}$ $8$ are $u = \sin^3(x)$ $8$ (for $u = \sin^3(x)$ $9$ ) and $x$ $0$ (for $x$ $1$ ).
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ . Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\newline$ $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$ Solving $\frac{dv}{dx}$ $4$ gives us $\frac{dv}{dx}$ $7$ within the interval $\frac{dv}{dx}$ $8$ . Solving $\frac{dv}{dx}$ $5$ , we can use the Pythagorean identity $u = \sin^3(x)$ $0$ to rewrite the equation as: $\newline$ $u = \sin^3(x)$ $1$ Simplifying the equation, we get: $\newline$ $u = \sin^3(x)$ $2$ $\newline$ $u = \sin^3(x)$ $3$ Taking the square root of both sides, we get: $\newline$ $u = \sin^3(x)$ $4$ The values of $x$ that satisfy $u = \sin^3(x)$ $4$ within the interval $\frac{dv}{dx}$ $8$ are $u = \sin^3(x)$ $8$ (for $u = \sin^3(x)$ $9$ ) and $x$ $0$ (for $x$ $1$ ). Now we have the $x$ -values of the turning points: $x$ $3$ and $x$ $4$ . To find the exact coordinates, we need to substitute these $x$ -values back into the original equation $x$ $6$ to find the corresponding $x$ $7$ -values.
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ . Substituting the derivatives into the product rule formula, we get: $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$ Solving $\frac{dv}{dx}$ $4$ gives us $\frac{dv}{dx}$ $7$ within the interval $\frac{dv}{dx}$ $8$ . Solving $\frac{dv}{dx}$ $5$ , we can use the Pythagorean identity $u = \sin^3(x)$ $0$ to rewrite the equation as: $u = \sin^3(x)$ $1$ Simplifying the equation, we get: $u = \sin^3(x)$ $2$ $u = \sin^3(x)$ $3$ Taking the square root of both sides, we get: $u = \sin^3(x)$ $4$ The values of $x$ that satisfy $u = \sin^3(x)$ $4$ within the interval $\frac{dv}{dx}$ $8$ are $u = \sin^3(x)$ $8$ (for $u = \sin^3(x)$ $9$ ) and $x$ $0$ (for $x$ $1$ ). Now we have the $x$ -values of the turning points: $x$ $3$ and $x$ $4$ . To find the exact coordinates, we need to substitute these $x$ -values back into the original equation $x$ $6$ to find the corresponding $x$ $7$ -values. Substituting $x$ $8$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $0$ . Substituting $3\sin^2(x)\cos(x)$ $1$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $3$ . Substituting $3\sin^2(x)\cos(x)$ $4$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $6$ . Substituting $3\sin^2(x)\cos(x)$ $7$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $0$ . Substituting $v = \cos(x)$ $0$ into $x$ $6$ , we get $v = \cos(x)$ $2$ . Substituting $v = \cos(x)$ $3$ into $x$ $6$ , we get $v = \cos(x)$ $5$ . Substituting $v = \cos(x)$ $6$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $0$ .
Calculate Turning Points: Now we need to find $\frac{du}{dx}$ and $\frac{dv}{dx}$ . The derivative of $u = \sin^3(x)$ with respect to $x$ is $3\sin^2(x)\cos(x)$ by using the chain rule. The derivative of $v = \cos(x)$ with respect to $x$ is $-\sin(x)$ .Substituting the derivatives into the product rule formula, we get: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos(x) \cdot \cos(x) + \sin^3(x) \cdot (-\sin(x))$ Simplifying the expression, we have: $\newline$ $\frac{dy}{dx} = 3\sin^2(x)\cos^2(x) - \sin^4(x)$ To find the critical points, we set the derivative equal to zero: $\newline$ $\frac{dv}{dx}$ $0$ We can factor out $\frac{dv}{dx}$ $1$ from the equation: $\newline$ $\frac{dv}{dx}$ $2$ Setting each factor equal to zero gives us the possible $x$ -values for the turning points: $\newline$ $\frac{dv}{dx}$ $4$ or $\frac{dv}{dx}$ $5$ Solving $\frac{dv}{dx}$ $4$ gives us $\frac{dv}{dx}$ $7$ within the interval $\frac{dv}{dx}$ $8$ .Solving $\frac{dv}{dx}$ $5$ , we can use the Pythagorean identity $u = \sin^3(x)$ $0$ to rewrite the equation as: $\newline$ $u = \sin^3(x)$ $1$ Simplifying the equation, we get: $\newline$ $u = \sin^3(x)$ $2$ $\newline$ $u = \sin^3(x)$ $3$ Taking the square root of both sides, we get: $\newline$ $u = \sin^3(x)$ $4$ The values of $x$ that satisfy $u = \sin^3(x)$ $4$ within the interval $\frac{dv}{dx}$ $8$ are $u = \sin^3(x)$ $8$ (for $u = \sin^3(x)$ $9$ ) and $x$ $0$ (for $x$ $1$ ).Now we have the $x$ -values of the turning points: $x$ $3$ and $x$ $4$ . To find the exact coordinates, we need to substitute these $x$ -values back into the original equation $x$ $6$ to find the corresponding $x$ $7$ -values.Substituting $x$ $8$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $0$ . $\newline$ Substituting $3\sin^2(x)\cos(x)$ $1$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $3$ . $\newline$ Substituting $3\sin^2(x)\cos(x)$ $4$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $6$ . $\newline$ Substituting $3\sin^2(x)\cos(x)$ $7$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $0$ . $\newline$ Substituting $v = \cos(x)$ $0$ into $x$ $6$ , we get $v = \cos(x)$ $2$ . $\newline$ Substituting $v = \cos(x)$ $3$ into $x$ $6$ , we get $v = \cos(x)$ $5$ . $\newline$ Substituting $v = \cos(x)$ $6$ into $x$ $6$ , we get $3\sin^2(x)\cos(x)$ $0$ .The exact coordinates of the turning points are therefore: $\newline$ $v = \cos(x)$ $9$ and $($2$\pi, $0$).

Find the exact coordinates of the turning points on the curve y=sin⁡3xcos⁡xy=\sin^3 x\cos xy=sin3xcosx for 000

Find the exact coordinates of the turning points on the curve $y=\sin^3 x\cos x$ for $0$