1.

Differentiate function with respect to x → d/dx (2x+3)⁶​

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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}


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