Question

What is the derivative wrt x of y = (3x² + 5)(log x - e)​

Answer 1

Solution :

$\textsf{Function of y =}\:\sf{(3x^{2} + 5)[In(x) - e]}$

$:\implies \sf{\dfrac{dy}{dx} = (3x^{2} + 5)[In(x) - e]}$

$\textsf{Now, by using the quotient rule of differentiation and substituting the values of v and u} \\ \textsf{ in it, we get :}$

$\underline{\sf{\bigstar\: Quotient\:rule\:of\: differentiation:-}} \\ \\ \sf{\dfrac{d(uv)}{dx} = (u)\cdot\dfrac{d(v)}{dx} + (v)\cdot\dfrac{d(u)}{dx}}$

$\sf{Here,} \\ \bullet\:\sf{The\:value\:of\:u = (3x^{2} + 5)} \\ \bullet\:\sf{The\:value\:of\:v = [In(x) - e]}$

$:\implies \sf{\dfrac{d(uv)}{dx} = (3x^{2} + 5)\cdot\dfrac{d[In(x) - e]}{dx} + [In(x) - e]\cdot\dfrac{d(3x^{2} + 5)}{dx}}$

$:\implies \sf{\dfrac{d(uv)}{dx} = (3x^{2} + 5)\cdot\dfrac{d[In(x)]}{dx} + \dfrac{d(-e)}{dx} + [In(x) - e]\cdot\dfrac{d(3x^{2})}{dx} + \dfrac{d(5)}{dx}}$

$\textsf{Now, by using the power rule of differentiation}\:\textsf{and constant rule of differentiation, we get :}$

$\underline{\sf{\bigstar\: Power\:rule\:of\: differentiation:-}} \\ \\ \sf{\dfrac{d(x^{n})}{dx} = n\cdot x^{(n - 1)}} \\ \\ \underline{\sf{\bigstar\: Constant\:rule\:of\: differentiation:-}} \\ \\ \sf{\dfrac{d(c)}{dx} = 0}$

$:\implies \sf{\dfrac{d(uv)}{dx} = (3x^{2} + 5)\cdot\bigg(\dfrac{1}{x} + 0\bigg) + [In(x) - e]\cdot(6x + 0)}$

$:\implies \sf{\dfrac{d(uv)}{dx} = \dfrac{(3x^{2} + 5)}{x} + 6x[In(x) - e]}$

$\boxed{\therefore \sf{\dfrac{dy}{dx} = \dfrac{(3x^{2} + 5)}{x} + 6x[In(x) - e]}}$