Question

Differentiate : X^3 + 3X^2 + X

Answer 1

Answer:

Given: y = x³ + 3x² + x

To find : Differentiation of "y" w.r.t "x"

i.e. dy/dx

Calculation :

y = x³ + 3x² + x

=> dy/dx = d(x³)/dx + 3 d(x²)/dx + d(x)/dx

=> dy/dx = 3x² + (3×2)x +1

=> dy/dx = 3x² + 6x + 1

Now, Another information about Differentiation :

1. It will be read as differential of "y" with respect to "x".

2. It represents the instantaneous change of "y" with change in "x".

3. Average change in "y" w.r.t to "x" is denoted as (∆y/∆x).

Answer 2

$\large{\underline{\underline{\mathfrak{\green{\sf{Solution:-}}}}}}$.

$\large{\underline{\underline{\mathfrak{\sf{Find\:here:-}}}}}$.

$\bold{\:Differentiate\:\:(x^3+3x^2+x)}$

$\large{\underline{\underline{\mathfrak{\sf{Explanation:-}}}}}$.

We can write this ,

$\leadsto\:Y\:=\:(x^3+3x^2+x)$

Differenciate W.R.To X ,

We find here ,

$\leadsto\frac{dY}{dX}\:=\frac{dy}{dx}\:(x^3+3x^2+x)$...........(1)

We know ,

• if $\:y\:=\:x^n$

So, Differentiate w.r. to x will be

• $\frac{dy}{dx}\:=\:n×\:x^{(n-1)}$

Now, by (1)

$\leadsto\boxed{\frac{dy}{dx}\:=\:(3x^2+6x+1)}$

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