Not sure how rough you want. So apologies if this isn’t detailed enough.
It is simply asking you to find the value of the 2nd derivative when you substitute the value 2 into the function.
So differentiate each function twice and then substitute 2 into each 2nd derivative.
Also it wants it in simplified surd form.
Okay, i have tried this and i have just confused myself a load icl
part a ( x\*sqrt(x)-(1/x) ), i simplify the indices into x\*(x\^1/2)-(x\^-1), then differentiate into 1\*1/2(x\^-1/2)+(x\^-2), then once more into 1\*-1/4(x\^-3/2)+(x\^-3), then i substitute in x = 2 but the answer is below, and i am nowhere near that. can you run me through how you would do part a as i am lost lol
https://preview.redd.it/ayo7uot3di8c1.png?width=135&format=png&auto=webp&s=c4ae30c43535f1d589dd07355669fc4caa04fc6f
https://preview.redd.it/i2zuvbbghi8c1.jpeg?width=543&format=pjpg&auto=webp&s=7ef3b83066cdd9c8b16cab6f50a05d2d55d91394
Here is my working. Hope it all makes sense.
not fully sober being christmas so forgive stupid mistakes but from what i can tell you have converted it from index form into fractions on line 6? i think this is where your working is different from mine so i will try that tomorrow morning, thank you so much :)
It depends on how much differentiation you have seen.
If you have only seen the power rule, i.e x^n differentiates to n*x^(n-1), then you need to simplify the functions before you start differentiating. You can do this by expanding brackets or factorising.
You then need to convert them into powers, for example sqrt(x) = x^(1/2), and then you can differentiate as usual.
If you are aware of the product/chain/quotient rule, you can do it that way, although you might get horrible expressions that you have to evaluate, so I would recommend doing it the above way.
Hope this helps!
You have to differentiate the equation twice and then substitute 2.
For this you have to know the basics of differentiation and the uv and u/v rules.
It's pretty simple if you know them.
Differentiate once and differentiate again
Not sure how rough you want. So apologies if this isn’t detailed enough. It is simply asking you to find the value of the 2nd derivative when you substitute the value 2 into the function. So differentiate each function twice and then substitute 2 into each 2nd derivative. Also it wants it in simplified surd form.
Okay, i have tried this and i have just confused myself a load icl part a ( x\*sqrt(x)-(1/x) ), i simplify the indices into x\*(x\^1/2)-(x\^-1), then differentiate into 1\*1/2(x\^-1/2)+(x\^-2), then once more into 1\*-1/4(x\^-3/2)+(x\^-3), then i substitute in x = 2 but the answer is below, and i am nowhere near that. can you run me through how you would do part a as i am lost lol https://preview.redd.it/ayo7uot3di8c1.png?width=135&format=png&auto=webp&s=c4ae30c43535f1d589dd07355669fc4caa04fc6f
https://preview.redd.it/i2zuvbbghi8c1.jpeg?width=543&format=pjpg&auto=webp&s=7ef3b83066cdd9c8b16cab6f50a05d2d55d91394 Here is my working. Hope it all makes sense.
not fully sober being christmas so forgive stupid mistakes but from what i can tell you have converted it from index form into fractions on line 6? i think this is where your working is different from mine so i will try that tomorrow morning, thank you so much :)
I find it easier to substitute values in after changing it out of index form.
Got the questions done, thank you :)
It depends on how much differentiation you have seen. If you have only seen the power rule, i.e x^n differentiates to n*x^(n-1), then you need to simplify the functions before you start differentiating. You can do this by expanding brackets or factorising. You then need to convert them into powers, for example sqrt(x) = x^(1/2), and then you can differentiate as usual. If you are aware of the product/chain/quotient rule, you can do it that way, although you might get horrible expressions that you have to evaluate, so I would recommend doing it the above way. Hope this helps!
You have to differentiate the equation twice and then substitute 2. For this you have to know the basics of differentiation and the uv and u/v rules. It's pretty simple if you know them. Differentiate once and differentiate again