f-x-2x-3-f-x-

Question Number 95339 by mathocean1 last updated on 24/May/20

f (x) =∣ 2 x + 3 ∣

f^{'} (x) = \dots ?

Answered by john santu last updated on 24/May/20

f (x) = \sqrt{{(2 x + 3)}^{2}}

f^{'} (x) = \frac{2 {(2 x + 3)}^{1} . 2}{2 \sqrt{{(2 x + 3)}^{2}}}

f^{'} (x) = \frac{4 x + 6}{∣ 2 x + 3 ∣}

Answered by prakash jain last updated on 24/May/20

x = \frac{y}{2} - \frac{3}{2} \Rightarrow y = 2 (x + \frac{3}{2})

∣ 2 x - 3 ∣=∣ y ∣

\frac{d f (x)}{d x} = \frac{d}{d y} ∣ y ∣ \times \frac{d y}{d x} = \frac{y}{∣ y ∣} \times 2

= \frac{2 (x + \frac{3}{2})}{∣ 2 (x + \frac{3}{2}) ∣} \times 2 = \frac{4 x + 6}{∣ 2 x + 3 ∣}

Commented by john santu last updated on 24/May/20

yes . your are right

Answered by M±th+et+s last updated on 24/May/20

b y f i r s t p r i n c i p l e

\underset{h \to 0}{l i m} \frac{∣ 2 (x + h) + 3 ∣ - ∣ 2 x + 3 ∣}{h} . \frac{∣ 2 (x + h) + 3 ∣ + ∣ 2 x + 3 ∣}{∣ 2 (x + h) + 3 ∣ + ∣ 2 x + 3 ∣}

\underset{h \to 0}{l i m} \frac{∣ (2 x + 3) + 2 h ∣^{2} - ∣ 2 x + 3 ∣^{2}}{h (∣ 2 (x + h) + 3 ∣ + ∣ 2 x + 3 ∣)}

\underset{h \to 0}{l i m} \frac{{(2 x + 3)}^{2} + 4 (2 x + 3) h + 4 h^{2} - {(2 x + 3)}^{2}}{h (∣ (2 x + 3) + 2 h ∣ + ∣ 2 x + 3 ∣)}

\underset{h \to 0}{l i m} \frac{4 h (2 x + 3 + h)}{h (∣ (2 x + 3) + 2 h ∣ + ∣ 2 x + 3 ∣)}

\underset{x \to 0}{l i m} \frac{4 (2 x + 3 + h)}{∣ 2 x + 3 + h ∣ + ∣ 2 x + 3 ∣} = \frac{4 x + 6}{∣ 2 x + 3 ∣}

Commented by john santu last updated on 24/May/20

∣ 2 (x + h) + 3 ∣ = ∣ (2 x + 3) + 2 h ∣

Commented by M±th+et+s last updated on 24/May/20

y e s i {d i d n}^{'} t f o c u s . t h a n k y o u s i r

Answered by mathmax by abdo last updated on 24/May/20

f (x) = {\begin{cases} 2 x + 3 if x > - \frac{3}{2} and f (- \frac{3}{2}) = 0 \Rightarrow \\ - 2 x - 3 if x < - \frac{3}{2} \end{cases}

f^{^{'}} (x) = {\begin{cases} 2 if x > - \frac{3}{2} \\ - 2 if x < - \frac{3}{2} \end{cases}

rest studying derivability at x_{0} = - \frac{3}{2} \dots .

Answered by 1549442205 last updated on 25/May/20

we have f^{2} (x) = {4 x}^{2} + 12 x + 9 \Rightarrow

2 f (x) . f^{^{'}} (x) = 8 x + 12 \Rightarrow f^{^{'}} (x) = \frac{8 x + 12}{2 f (x)} = \frac{4 x + 6}{∣ 2 x + 3 ∣}

Answered by mathmax by abdo last updated on 10/Oct/21

f (x) = \sqrt{{(2 x + 3)}^{2}} \Rightarrow f^{^{'}} (x) = \frac{2.2 (2 x + 3)}{2 \sqrt{{(2 x + 3)}^{2}}} = \frac{4 x + 6}{∣ 2 x + 3 ∣}

Commented by prakash jain last updated on 10/Oct/21

how are you, sir