Weird Differentiation

Calculus Level 3

Let us define a new differential operator as follows:

d ~ d ~ x f ( x ) = lim ⁡ h → 0 f ( x + h ) f ′ ( x + h ) − f ( x ) f ′ ( x ) h \frac {\tilde d}{\tilde dx} f(x) = \lim_{h \to 0} \frac {f(x+h)f'(x+h)-f(x)f'(x)}h

where f ′ ( x ) = d d x f ( x ) f'(x) = \dfrac d{dx} f(x) . Find d ~ d ~ x x 2 \dfrac {\tilde d}{\tilde dx} x^2 at x = 5 x=5 .


The answer is 150.

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1 solution

Chew-Seong Cheong
Mar 20, 2018

d ~ d ~ x x 2 = lim ⁡ h → 0 ( x + h ) 2 ⋅ d d x ( x + h ) 2 − x 2 ⋅ d d x x 2 h = lim ⁡ h → 0 ( x 2 + 2 h x + h 2 ) ⋅ 2 ( x + h ) − x 2 ⋅ 2 x h = lim ⁡ h → 0 2 ( x 3 + 3 h x 2 + 3 h 2 x + h 3 ) − 2 x 3 h = lim ⁡ h → 0 ( 6 x 2 + 6 h x + 2 h 2 ) = 6 x 2 \begin{aligned}\frac {\tilde d}{\tilde dx} x^2 & = \lim_{h \to 0} \frac {(x+h)^2\cdot \frac d{dx} (x+h)^2 - x^2 \cdot \frac d{dx}x^2}h \\ & = \lim_{h \to 0} \frac {\left(x^2+2hx+h^2\right)\cdot 2(x+h) - x^2 \cdot 2x}h \\ & = \lim_{h \to 0} \frac {2\left(x^3+3hx^2+3h^2x+h^3\right) - 2x^3}h \\ & = \lim_{h \to 0} \left(6x^2 + 6hx + 2h^2\right) \\ & = 6x^2 \end{aligned}

Therefore, d ~ d ~ x x 2 ∣ x = 5 = 6 ⋅ 5 2 = 150 \dfrac {\tilde d}{\tilde dx} x^2 \bigg|_{x=5} = 6\cdot 5^2 = \boxed{150} .

5 Helpful 0 Interesting 0 Brilliant 0 Confused

@Ayush Mishra , I redo the problem LaTex and wording for you.

Chew-Seong Cheong - 3 years, 2 months ago

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