Question

Trouvez la différentielle de la fonction y=x3⋅sinx

Answer 1

The answer is "
`y'= x^2( x \cos x + 3 \sin x)`"

Explanation:

Given:

`\to y= x^3 \cdot \sin x\\\\`

Formula:

`\to (d)/(dx) (a b) = a (db)/(dx)+b (da)/(dx)`

`\to (d)/(dx) x^n = nx^(n-1)\\\\\to (d)/(dx) \sin x = \cos x\\`

Differentiate the given value:

`\to y= x^3 \cdot \sin x\\\\ \to y' = x^3 (d)/(dx) (\sin x) + \sin x (d)/(dx) (x^3)\\\\ \to y'= x^3 \cos x + \sin x \cdot 3x^2\\\\\to y'= x^2( x \cos x + 3 \sin x) \\\\`