$T(x) = \frac{e^{x^2 + 8x}}{\sqrt{x^4 + 7}},\quad T'(x) =$

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Find higher derivatives of rational and radical functions

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T(x) = \frac{e^{x^2 + 8x}}{\sqrt{x^4 + 7}},\quad T'(x) =

Apply Quotient Rule: To find the derivative of the function $T(x) = \frac{e^{x^2 + 8x}}{\sqrt{x^4 + 7}}$ , we will use the quotient rule and the chain rule. The quotient rule states that the derivative of a function that is the quotient of two other functions, $\frac{u(x)}{v(x)}$ , is given by $\frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$ . The chain rule allows us to differentiate composite functions.
Identify $u(x)$ and $v(x)$ : Let's identify $u(x)$ and $v(x)$ for the quotient rule. We have: $\newline$ $u(x) = e^{(x^2 + 8x)}$ $\newline$ $v(x) = \sqrt{x^4 + 7}$ $\newline$ Now we need to find the derivatives $u'(x)$ and $v'(x)$ .
Find $u'(x)$ : First, we find $u'(x)$ using the chain rule. The derivative of $e^{g(x)}$ with respect to $x$ is $e^{g(x)}g'(x)$ , where $g(x) = x^2 + 8x$ . So we need to find the derivative of $g(x)$ , which is $g'(x) = 2x + 8$ .
Find $v'(x)$ : Now we calculate $u'(x)$ :
$u'(x) = e^{(x^2 + 8x)} \cdot (2x + 8)$
Apply Chain Rule: Next, we find $v'(x)$ . Since $v(x) = \sqrt{x^4 + 7}$ , we can rewrite this as $(x^4 + 7)^{\frac{1}{2}}$ . Using the chain rule, the derivative of $h(x)^n$ is $n \cdot h(x)^{n-1} \cdot h'(x)$ , where $h(x) = x^4 + 7$ and $n = \frac{1}{2}$ .
Calculate $u'(x)$ : We calculate the derivative of $h(x)$ , which is $h'(x) = 4x^3$ . Then we apply the chain rule to find $v'(x)$ : $v'(x) = \frac{1}{2} \cdot (x^4 + 7)^{-\frac{1}{2}} \cdot 4x^3$
Apply Quotient Rule: Simplify $v'(x)$ : $v'(x) = \frac{2x^3}{\sqrt{x^4 + 7}}$
Substitute into Formula: Now we apply the quotient rule to find $T'(x)$ : $T'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$
Simplify Expression: Substitute $u(x)$ , $u'(x)$ , $v(x)$ , and $v'(x)$ into the quotient rule formula: $\newline$ $T'(x) = \frac{\sqrt{x^4 + 7} \cdot e^{x^2 + 8x} \cdot (2x + 8) - e^{x^2 + 8x} \cdot 2x^3}{\sqrt{x^4 + 7}} / (x^4 + 7)$
Final Answer: Simplify the expression by combining terms and factoring out common factors where possible: $\newline$ $T'(x) = \frac{e^{(x^2 + 8x)} \cdot ((2x + 8) \cdot \sqrt{x^4 + 7} - 2x^3)}{x^4 + 7}$
Final Answer: Simplify the expression by combining terms and factoring out common factors where possible: $\newline$ $T'(x) = \frac{e^{x^2 + 8x} \cdot ((2x + 8) \cdot \sqrt{x^4 + 7} - 2x^3)}{x^4 + 7}$ We have found the derivative of $T(x)$ with respect to $x$ . The final answer is: $\newline$ $T'(x) = \frac{e^{x^2 + 8x} \cdot ((2x + 8) \cdot \sqrt{x^4 + 7} - 2x^3)}{x^4 + 7}$