Question

Find the derivative of the function.
h(x) = 7x^2 − x

Answer 1

\[Thats 7x ^{2-x}\]

Answer 2

14x-1

Answer 3

oh nvm lol use chain rule

Answer 4

Is it ln(7)(7)\[x^{2-x}\]

Answer 5

include x^2-x after 7

Answer 6

i believe its -x^(1-x)

Answer 7

@Spidermonkeyy im pretty sure it should look similar to \[\ln(7)(7)^{x ^{2}-x}\]

Answer 8

because thats what all the examples look like

Answer 9

i used chain rule but im not sure about lns in there sorry

Answer 10

not h(x) = 7x^2 -x
its
\[7x^{2-x}\]

Answer 11

Ok

Answer 12

\[h(x) = 7x^{2 -x} \]
taking log both sides we find:
\[\log h(x) = \log(7x^{2 -x}) \rightarrow \log h(x) = \log7 + \log x^{2 -x}\]
\[\rightarrow \log h(x) = \log7 +(2-x) \log x\]
Differentiating with respect to x we find
\[\frac{1}{h(x)} h'(x) = 0 + (2-x) \frac{1}{x} + \log x (0-1)\]
\[\rightarrow \frac{1}{h(x)} h'(x) = \frac{(2-x)}{x} + \log x (-1)\]
\[\rightarrow h'(x) = h(x) (\frac{(2-x)}{x} - \log x)\]
\[\rightarrow h'(x) = 7x^{2-x} (\frac{(2-x)}{x} - \log x)\]
\[\rightarrow h'(x) = 7x^{2-x} (\frac{(2-x-x \log x)}{x} )\]
\[\rightarrow h'(x) = 7x^{2-1-x} (2-x-x \log x)\]
\[\rightarrow h'(x) = 7x^{1-x} (2-x-x \log x)\]

@arshia93

Answer 13

@arshia93 Your question has been solved.....

Answer 14

@dpasingh bravo! Can you please answer a different question I have if you cant find it ill bump it in 7 mins