Question

differentiate (ax^2+bx+c) (x-d)
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Answer 1

Given:

An expression (ax²+bx+c)(x-d)

To Find:

Differentiation of given expression

Solution:

We know that,

$\longrightarrow\rm{\dfrac{d}{dx}(x^{n})=nx^{n-1}}$

$\longrightarrow\rm{\dfrac{d}{dx}(constant)=0}$

$\rule{190}{1}$

Let y= (ax²+bx+c)(x-d)

So,

y= x(ax²+bx+c)-d(ax²+bx+c)

y= ax³+bx²+cx-adx²-bdx-cd

y= ax³+(b-ad)x²+(c-bd)x-cd

Here, a,b,c and d are constants

So, on differentiating y w.r.t. x, we get

$\rightarrow\sf{\dfrac{dy}{dx}=\dfrac{d}{dx}(ax^{3}+(b-ad)x^{2}+(c-bd)x-cd)}$

$\longrightarrow\sf{\dfrac{dy}{dx}=3ax^{2}+2(b-ad)x+(c-bd)-0}$

$\longrightarrow\sf\green{\dfrac{dy}{dx}=3ax^{2}+2(b-ad)x+c-bd}$

Hence, differentiation of given expression is 3ax²+2(b-ad)x+c-bd.

Formulae to Remember

$\longrightarrow\rm{\dfrac{d}{dx}(ln\:x)=\dfrac{1}{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(e^{x})=e^{x}}$

$\longrightarrow\rm{\dfrac{d}{dx}(f(x))=f'(x)}$

$\longrightarrow\rm{\dfrac{d}{dx}(sinx)=cosx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosx)=-sinx}$

$\longrightarrow\rm{\dfrac{d}{dx}(tanx)=sec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(cotx)=-cosec^{2}x}$

$\longrightarrow\rm{\dfrac{d}{dx}(secx)=secx.tanx}$

$\longrightarrow\rm{\dfrac{d}{dx}(cosecx)=-cosecx.cotx}$

Answer 2

GIVEN EXPRESSION :-

(ax² + bx + c)(x - d)

OBJECTIVE :-

To differentiate the given expression.

PRE-REQUISITE :-

Let y = ax² + bx + c.

SOLUTION :-

$\sf{\dfrac{dy}{dx}=\dfrac{d(ax^2+bx+c)}{dx}}\\\\\sf{=\dfrac{d[(x-d)(ax^2+(ax^2+bx+c))]}{dx}} \\\\\sf{By\ product\ rule:-}\\\\\sf{=\dfrac{d(x-d).(ax^2+bx+c)}{dx}+(x-d).\dfrac{d(ax^2+bx+c)}{dx}}\\\\\sf{=[\dfrac{d}{dx}(x)+\dfrac{d}{dx}(-x)](ax^2+bx+c)(x-d)[a.\dfrac{d}{dx}(x^2)+b.\dfrac{d}{dx}(x)+\dfrac{d}{dx}(c)] }\\\\\sf{=(1+0)(ax^2+bx+c)+(x-d)(a.2x+b.1+0)}\\\\\boxed{\sf{=ax^2+(x-d)(2ax+b)+bx+c}}$

$\rule{200}{1.9}$

• Product rule :- [a(x) . b(x)]' = a'(x) + b(x) + a(x) + b'(x)

• Derivative of the differentiation of a variable is 1.

• Derivative of the differentiation of a constant is 0.

• Power rule :- (xⁿ)' = n . xⁿ⁻¹