Question

if

y = x^4+x^2+1

then ,

dy/dx = ????

Answer 1

Given,

y = x ^ 4 + x² + 1



Differentiate both sides with respect to x

dy/dx = d(x^4 + x² + 1 ) / dx

dy/dx =d(x^4)/dx + d(x²)/dx + d(1)/dx

dy/dx = 4x³ + 2x + 0

dy/dx = 4x³ + 2x

Therefore, dy/dx = 4x³ + 2x

We used :

d/dx ( u + v) = du/dx + dv/dx

d/dx( x^n ) = n * x ^ n - 1

d/dx ( k) = 0 , where k is constant.

d/dx ( kx^n ) = k * d/dx ( x^n) , where k is constant.

Answer 2

Hi friend, Harish here.

Here is your answer.

$y = x^4 + x^2 + 1 \\ \\ \frac{dy}{dx} = \frac{d}{dx}\bigl( x^4 +x^2 +1 \bigr ) \\ \\ Now, Apply\ sum \ rule \\ \\ Then,\\ \\ \frac{d}{dx}(y) = \frac{d}{dx}(x^4) + \frac{d}{dx}(x^2)+\frac{d}{dx}(1) \\ \\ \to \bigl (4.x^{(4-1)} \bigr ) + \bigl( 2. x^{(2-1)} \bigr ) + 0 . \\ \\ \to 4x^3 + 2x \\ \\ \boxed{\bold{Therefore\ \frac{dy}{dx} = 4x^3 + 2x }}$
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Hope my answer is helpful to you.