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Finding the derivative
- Thread starter Khan101
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Do you know - How to find the derivative of \(\displaystyle \sqrt{x+1}\) using chain-rule ?Hey all,
Wondering if anyone could provide working out to answer the following question:
--> Derive 6x(√x+1).... NOTE: x+1 is within square root.
Cheers,
Yes. That equals to[math]1\div2\sqrt{\smash[b]{x}}[/math]Correct me if I am wrong?Do you know - How to find the derivative of \(\displaystyle \sqrt{x+1}\) using chain-rule ?
But how do I include '6x' into this problem?
Subhotosh,Do you know - How to find the derivative of \(\displaystyle \sqrt{x+1}\) using chain-rule ?
Hey all,
Wondering if anyone could provide working out to answer the following question:
--> Derive 6x(√x+1).... NOTE: x+1 is within square root.
Cheers,
UPDATE!!!
I have derived up until this point: 6*sqrt(x+1)+(3*x)/sqrt(x+1)
How can I simplify/rewrite the equation to equal, (9*x+6)/sqrt(x+1)?
Thank you.
--> Derive 6x(√x+1).... NOTE: x+1 is within square root.
Just write the open parenthesis immediately after the radical symbol. A pair of parentheses
are not required around the radical expression here:
6x√(x+1)
__________________________________________________________________________________________________
You were asked if you could find the derivative of \(\displaystyle \ \sqrt{x + 1}.\)
You answered:
[math] \ 1\div2\sqrt{\smash[b]{x}}.[/math]
That is not correct. It would be[math]1\div(2\sqrt{\smash[b]{x + 1}}).[/math]
Or, not using that division sign, you might write it as \(\displaystyle \ \ \dfrac{1}{2\sqrt{x + 1}}.\)
Last edited:
Ahhhhh I see, thanks.Just write the open parenthesis immediately after the radical symbol. A pair of parentheses
are not required around the radical expression here:
6x√(x+1)
__________________________________________________________________________________________________
You were asked if you could find the derivative of \(\displaystyle \ \sqrt{x + 1}.\)
You answered:
[math] \ 1\div2\sqrt{\smash[b]{x}}.[/math]
That is not correct. It would be[math]1\div(2\sqrt{\smash[b]{x + 1}}).[/math]
Or, not using that division sign, you might write it as \(\displaystyle \ \ \dfrac{1}{2\sqrt{x + 1}}.\)
Always have to lookagain hahahahaahaha.
Last edited:
Thank you for your input Jeff.[math]y = 6x \sqrt{x + 1} \text { for } x > - 1 \implies\\ y’ = 6 \sqrt{x + 1} + \dfrac{6x}{2\sqrt{x + 1}} = \dfrac{6 \sqrt{x + 1} * \sqrt{x + 1}}{\sqrt{x + 1}}+ \dfrac{3x}{\sqrt{x + 1}} = \text {WHAT?}[/math]
The derivative of f(x)= [math]6x\sqrt{\smash[b]{x+1}}[/math] is equal to f'(x) = [math]9x+6\div\sqrt{\smash[b]{x+1}}[/math]
Cheers,
Not quite. Parentheses are important. \(\displaystyle f'(x) = (9x +6) \div \sqrt{x + 1}\)Thank you for your input Jeff.
The derivative of f(x)= [math]6x\sqrt{\smash[b]{x+1}}[/math] is equal to f'(x) = [math]9x+6\div\sqrt{\smash[b]{x+1}}[/math]
Cheers,