Given the function f(x)=(-4-6x^(3))(-9x^(3)-4x^(2)+5), find f^(')(x) in any form. Answer: f^(')(x)=

Apply Product Rule: To find the derivative of the function f(x) = (-4-6x^(3))(-9x^(3)-4x^(2)+5), we will use the product 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.Find Derivatives: Let's denote the first function as u(x) = -4-6x^(3) and the second function as v(x) = -9x^(3)-4x^(2)+5. We need to find the derivatives u'(x) and v'(x).Distribute Terms: First, we find the derivative of u(x) = -4-6x^(3). The derivative of a constant is 0, and the derivative of -6x^(3) is -18x^(2). So, u'(x) = 0 - 18x^(2) = -18x^(2).Simplify Expressions: Next, we find the derivative of v(x) = -9x^(3)-4x^(2)+5. The derivative of -9x^(3) is -27x^(2), the derivative of -4x^(2) is -8x, and the derivative of a constant is 0. So, v'(x) = -27x^(2) - 8x.Combine Like Terms: Now we apply the product rule: f'(x) = u'(x)v(x) + u(x)v'(x). Substituting the derivatives we found, we get f'(x) = (-18x^(2))(-9x^(3)-4x^(2)+5) + (-4-6x^(3))(-27x^(2) - 8x).We need to distribute the terms in the expression f'(x) = (-18x^(2))(-9x^(3)-4x^(2)+5) + (-4-6x^(3))(-27x^(2) - 8x). Let's do this step by step.First, distribute -18x^(2) across the terms in the first parenthesis: f'(x) = (-18x^(2))(-9x^(3)) + (-18x^(2))(-4x^(2)) + (-18x^(2))(5).Now, distribute -4 and -6x^(3) across the terms in the second parenthesis: f'(x) = f'(x) + (-4)(-27x^(2)) + (-4)(-8x) + (-6x^(3))(-27x^(2)) + (-6x^(3))(-8x).Let's simplify the terms we have so far: f'(x) = 162x^(5) + 72x^(4) - 90x^(2) + 108x^(2) + 32x - 162x^(5) - 48x^(4).Combine like terms to get the final derivative: f'(x) = 162x^(5) - 162x^(5) + 72x^(4) - 48x^(4) - 90x^(2) + 108x^(2) + 32x.After combining like terms, we get f'(x) = 24x^(4) + 18x^(2) + 32x.

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Given the function
f(x)=(-4-6x^(3))(-9x^(3)-4x^(2)+5), find
f^(')(x) in any form.
Answer:
f^(')(x)=

Given the function $f(x)=\left(-4-6 x^{3}\right)\left(-9 x^{3}-4 x^{2}+5\right)$ , find $f^{\prime}(x)$ in any form. $\newline$ Answer: $f^{\prime}(x)=$

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Calculus

Find derivatives of using multiple formulae

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Q. Given the function

f(x)=\left(-4-6 x^{3}\right)\left(-9 x^{3}-4 x^{2}+5\right)

, find

f^{\prime}(x)

in any form.

\newline

Answer:

f^{\prime}(x)=

Apply Product Rule: To find the derivative of the function $f(x) = (-4-6x^{3})(-9x^{3}-4x^{2}+5)$ , we will use the product 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.
Find Derivatives: Let's denote the first function as $u(x) = -4-6x^{3}$ and the second function as $v(x) = -9x^{3}-4x^{2}+5$ . We need to find the derivatives $u'(x)$ and $v'(x)$ .
Distribute Terms: First, we find the derivative of $u(x) = -4-6x^{3}$ . The derivative of a constant is $0$ , and the derivative of $-6x^{3}$ is $-18x^{2}$ . So, $u'(x) = 0 - 18x^{2} = -18x^{2}$ .
Simplify Expressions: Next, we find the derivative of $v(x) = -9x^{3}-4x^{2}+5$ . The derivative of $-9x^{3}$ is $-27x^{2}$ , the derivative of $-4x^{2}$ is $-8x$ , and the derivative of a constant is $0$ . So, $v'(x) = -27x^{2} - 8x$ .
Combine Like Terms: Now we apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$ . Substituting the derivatives we found, we get $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ .
Combine Like Terms: Now we apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$ . Substituting the derivatives we found, we get $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ .We need to distribute the terms in the expression $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ . Let's do this step by step.
Combine Like Terms: Now we apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$ . Substituting the derivatives we found, we get $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ .We need to distribute the terms in the expression $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ . Let's do this step by step.First, distribute $-18x^{2}$ across the terms in the first parenthesis: $f'(x) = (-18x^{2})(-9x^{3}) + (-18x^{2})(-4x^{2}) + (-18x^{2})(5)$ .
Combine Like Terms: Now we apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$ . Substituting the derivatives we found, we get $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ .We need to distribute the terms in the expression $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ . Let's do this step by step.First, distribute $-18x^{2}$ across the terms in the first parenthesis: $f'(x) = (-18x^{2})(-9x^{3}) + (-18x^{2})(-4x^{2}) + (-18x^{2})(5)$ .Now, distribute $-4$ and $-6x^{3}$ across the terms in the second parenthesis: $f'(x) = f'(x) + (-4)(-27x^{2}) + (-4)(-8x) + (-6x^{3})(-27x^{2}) + (-6x^{3})(-8x)$ .
Combine Like Terms: Now we apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$ . Substituting the derivatives we found, we get $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ .We need to distribute the terms in the expression $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ . Let's do this step by step.First, distribute $-18x^{2}$ across the terms in the first parenthesis: $f'(x) = (-18x^{2})(-9x^{3}) + (-18x^{2})(-4x^{2}) + (-18x^{2})(5)$ .Now, distribute $-4$ and $-6x^{3}$ across the terms in the second parenthesis: $f'(x) = f'(x) + (-4)(-27x^{2}) + (-4)(-8x) + (-6x^{3})(-27x^{2}) + (-6x^{3})(-8x)$ .Let's simplify the terms we have so far: $f'(x) = 162x^{5} + 72x^{4} - 90x^{2} + 108x^{2} + 32x - 162x^{5} - 48x^{4}$ .
Combine Like Terms: Now we apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$ . Substituting the derivatives we found, we get $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ .We need to distribute the terms in the expression $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ . Let's do this step by step.First, distribute $-18x^{2}$ across the terms in the first parenthesis: $f'(x) = (-18x^{2})(-9x^{3}) + (-18x^{2})(-4x^{2}) + (-18x^{2})(5)$ .Now, distribute $-4$ and $-6x^{3}$ across the terms in the second parenthesis: $f'(x) = f'(x) + (-4)(-27x^{2}) + (-4)(-8x) + (-6x^{3})(-27x^{2}) + (-6x^{3})(-8x)$ .Let's simplify the terms we have so far: $f'(x) = 162x^{5} + 72x^{4} - 90x^{2} + 108x^{2} + 32x - 162x^{5} - 48x^{4}$ .Combine like terms to get the final derivative: $f'(x) = 162x^{5} - 162x^{5} + 72x^{4} - 48x^{4} - 90x^{2} + 108x^{2} + 32x$ .
Combine Like Terms: Now we apply the product rule: $f'(x) = u'(x)v(x) + u(x)v'(x)$ . Substituting the derivatives we found, we get $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ .We need to distribute the terms in the expression $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ . Let's do this step by step.First, distribute $-18x^{2}$ across the terms in the first parenthesis: $f'(x) = (-18x^{2})(-9x^{3}) + (-18x^{2})(-4x^{2}) + (-18x^{2})(5)$ .Now, distribute $-4$ and $-6x^{3}$ across the terms in the second parenthesis: $f'(x) = f'(x) + (-4)(-27x^{2}) + (-4)(-8x) + (-6x^{3})(-27x^{2}) + (-6x^{3})(-8x)$ .Let's simplify the terms we have so far: $f'(x) = 162x^{5} + 72x^{4} - 90x^{2} + 108x^{2} + 32x - 162x^{5} - 48x^{4}$ .Combine like terms to get the final derivative: $f'(x) = 162x^{5} - 162x^{5} + 72x^{4} - 48x^{4} - 90x^{2} + 108x^{2} + 32x$ .After combining like terms, we get $f'(x) = (-18x^{2})(-9x^{3}-4x^{2}+5) + (-4-6x^{3})(-27x^{2} - 8x)$ $0$ .

Given the function $f(x)=\left(-4-6 x^{3}\right)\left(-9 x^{3}-4 x^{2}+5\right)$ , find $f^{\prime}(x)$ in any form. $\newline$ Answer: $f^{\prime}(x)=$

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Given the function f(x)=(−4−6x3)(−9x3−4x2+5) f(x)=\left(-4-6 x^{3}\right)\left(-9 x^{3}-4 x^{2}+5\right) f(x)=(−4−6x3)(−9x3−4x2+5), find f′(x) f^{\prime}(x) f′(x) in any form.\newlineAnswer: f′(x)= f^{\prime}(x)= f′(x)=

Given the function $f(x)=\left(-4-6 x^{3}\right)\left(-9 x^{3}-4 x^{2}+5\right)$ , find $f^{\prime}(x)$ in any form. $\newline$ Answer: $f^{\prime}(x)=$