Quick math question? (1 Viewer)

[Zokunu]

Differentiate f(x) = √x+2 from first principle.

Ok, here is what I got at the end...

1/h x h/√(x+h)+2 + √x+2

The answer is 1/2√x+2 . But how?

Thanks.

 

[Carrotsticks]

You can now let h -> 0, after cancelling some of them.

 

[Zokunu]

You can now let h -> 0, after cancelling some of them.

this?

1/√(x+0)+2 +Vx+2 ?

 

[Carrotsticks]

You are missing a lot of brackets, which makes it confusing. Is this your expression?

 

[Zokunu]

You are missing a lot of brackets, which makes it confusing. Is this your expression?

mhm, ye

 

[Carrotsticks]

Remember, differentiation by first principles has a lim as h->0 at the front. If we cancel the H in the denominator and numerator, and then let H->0, we get the required expression.

 

[Zokunu]

Remember, differentiation by first principles has a lim as h->0 at the front. If we cancel the H in the denominator and numerator, and then let H->0, we get the required expression.

1/√x+2 + √x+2
1/2√x+2

Thanks man

 

[Menomaths]

Edit:Yis I'm becoming good at latex

 

[Zokunu]

Edit:Yis I'm becoming good at latex

thx Meno . One more question related to this. Find dy/dx if y= a^2 + b^3 + c

isn't supposed to be 2a + 3b^2?
The answer is 0?

 

[Menomaths]

thx Meno . One more question related to this. Find dy/dx if y= a^2 + b^3 + c

isn't supposed to be 2a + 3b^2?
The answer is 0?

I think the answer is 0 because you have to differentiate with respect to y. a,b, and c are basically constants

 

[Carrotsticks]

I think the answer is 0 because you have to differentiate with respect to y. a,b, and c are basically constants

Yep, this.

Suppose y=x^3, then dy/dx = 3x^2 as we usually do because we are differentiating with respect to x.

However, if we were to differentiate with respect to z, we would have dy/dz = 0, since in the 'eyes of the z function', y=x^3 is just a Constant function.

He has no z's in him, so therefore 'dy/dz' is not interested in him and treat him as a Constant.

Nobody cares about Constant.

Hence, dy/dz = 0.

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Alternatively, y=x^3 = y=z^0 * x^3, and differentiating with respect to z makes us 'bring the 0 down', hence we get 0.