{:[T(x)=(e^(x^(2)+8x))/(sqrt(x^(4)+7))],[T^(')(x)=]:}

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Apply Quotient Rule: To find the derivative of the function T(x) = (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, u(x)/v(x), is given by (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: u(x) = e^(x^2 + 8x) v(x) = sqrt(x^4 + 7) 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) * (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)^(1/2). Using the chain rule, the derivative of h(x)^n is n*h(x)^(n-1)*h'(x), where h(x) = x^4 + 7 and n = 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) = (1/2) * (x^4 + 7)^(-1/2) * 4x^3Apply Quotient Rule: Simplify v'(x): v'(x) = 2x^3 / sqrt(x^4 + 7)Substitute into Formula: Now we apply the quotient rule to find T'(x): T'(x) = (v(x)u'(x) - u(x)v'(x)) / (v(x))^2Simplify Expression: Substitute u(x), u'(x), v(x), and v'(x) into the quotient rule formula: T'(x) = (sqrt(x^4 + 7) * e^(x^2 + 8x) * (2x + 8) - e^(x^2 + 8x) * 2x^3 / sqrt(x^4 + 7)) / (x^4 + 7)Final Answer: Simplify the expression by combining terms and factoring out common factors where possible: T'(x) = (e^(x^2 + 8x) * ((2x + 8) * 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: T'(x) = (e^(x^2 + 8x) * ((2x + 8) * sqrt(x^4 + 7) - 2x^3)) / (x^4 + 7)

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

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T(x)=ex2+8xx4+7,T′(x)=T(x) = \frac{e^{x^2 + 8x}}{\sqrt{x^4 + 7}},\quad T'(x) = T(x)=x4+7​ex2+8x​,T′(x)=

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