Mathematics, 27.07.2021 01:00, igtguith
Hello again! This is another Calculus question to be explained.
The prompt reads that "If f(x) is a twice-differentiable function such that f(2) = 2 and 
= 
, then what is the value of 
at x = 2?"
My initial calculation lead to 12, but then I guessed 219 as the answer and it was correct. Would any kind soul please explain why the answer would be 219? Thank you so much!
Answers: 2
Exponential Property [Root Rewrite]: ![\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}](/tpl/images/1400/0329/77a1d.png)
![\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)](/tpl/images/1400/0329/4bc1d.png)
![\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]](/tpl/images/1400/0329/9c114.png)
![\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)](/tpl/images/1400/0329/fb0e0.png)
![\displaystyle \frac{d^2y}{dx^2} = \frac{d}{dx} \Big[ 6(x^2 + 3y^2)^\big{\frac{1}{2}} \Big] \cdot \frac{d}{dx} \Big[ (x^2 + 3y^2) \Big]](/tpl/images/1400/0329/c349c.png)
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 6y(6\sqrt{x^2 + 3y^2}) \big]](/tpl/images/1400/0329/a4f51.png)
![\displaystyle \frac{d^2y}{dx^2} = 3(x^2 + 3y^2)^\big{\frac{-1}{2}} \big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]](/tpl/images/1400/0329/a0438.png)
![\displaystyle \frac{d^2y}{dx^2} = \frac{3\big[ 2x + 36y\sqrt{x^2 + 3y^2} \big]}{\sqrt{x^2 + 3y^2}}](/tpl/images/1400/0329/e3ae7.png)
![\displaystyle \frac{d^2y}{dx^2} \bigg| \limits_{x = 2} = \frac{3\big[ 2(2) + 36(2)\sqrt{2^2 + 3(2)^2} \big]}{\sqrt{2^2 + 3(2)^2}}](/tpl/images/1400/0329/491c5.png)
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Hello again! This is another Calculus question to be explained.
The prompt reads that "If f(x) is a...
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