Question

Find dy/dx for the following function
Y= x^3-3x^2+3x-2/5

Answer 1

Explanation:

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Answer 2

Given:

A function

y= x³-3x²+3x-2/5

To Find:

dy/dx of given function

Solution:

On differentiating y w.r.t x, we get

$\longrightarrow\rm{\dfrac{dy}{dx}=x^{3}-3x^{2}+3x-\dfrac{2}{5} }$

$\longrightarrow\rm{\dfrac{dy}{dx}=3x^{2}-3(2x)+3(1)-0}$

$\longrightarrow\rm\green{\dfrac{dy}{dx}=3x^{2}-6x+3}$

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Formulae to Remember-

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

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

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

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

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

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

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

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

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

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

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