Question

Differentiate function with respect to x →

d/dx (2x+3)⁶​

Answer 1

Answer:

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☞Differentiate function with respect to x →

❥$\dfrac{d( {2x + 3)}^{6} }{dx}$

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Given:-

• y=(2x+3)⁶

To Find:-

• $\dfrac{dy}{dx}$

Solution:-

Let (2x+3) be equals to t

t=2x+3

=>(2x+3)⁶=t⁶

• So We need to differentiate t⁶ with respect to x where t=2x+3

$= > \dfrac{d( {t}^{6}) }{dx} \\ \\ = \frac{d( {t}^{6}) }{dt} \times \frac{dt}{dx}$

(Multiplying and dividing dt in equation)

$= (6)( {t}^{6 - 1} ) \times \dfrac{dt}{dx}$

(Since, $\frac{d {x}^{n} }{dx} = n {x}^{n - 1}$)

$= 6 {t}^{5} \times \dfrac{dt}{dx} \\ = 6( {2x + 3})^{5} \times \frac{d(2x + 3)}{dx}$

(Putting value of t=2x+3)

$= 6( {2x + 3})^{5} \times (2 \times 1)( {x}^{1 - 1} ) \\ = 6 \times 2 \times ( {2x + 3)}^{5} \\ = 12 \times ({2x + 3})^{5}$

Therefore,Differentiating the function $\dfrac{d( {2x + 3)}^{6} }{dx}$ with respect to x

We get Solution as, 12(2x+3)⁵

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❥Basic Differentiation Formulae:-

• y=constant, $= > \frac{dy}{dx} = 0$

• y=xⁿ, $=>\frac{dy }{dx} = n {x}^{n - 1}$

• y=sinx, $=>\frac{dy }{dx} =cosx$

• y=cosx, $=>\frac{dy }{dx} =(-sinx)$

• y=tanx, $=>\frac{dy }{dx} =({ sec}^{2}x)$

• y=cotx,$= > \frac{dy}{dx} = { - cosec}^{2} x$

• $y = {a}^{x} = > \frac{dy}{dx} = {a}^{x} ln(a)$

• $y = {e}^{x} = > \frac{dy}{dx} = {e}^{x}$

• $y = ln(x ) = > \frac{dy}{dx} = \frac{1}{x}$