Question

differentiate with respect to x

Answer 1

Answer:

let fx=t

differentiation of t⁵ w.r.t. x is 5t⁴

then differentiation of t w.r.t. x

ans:

5(x³-2x-1)⁴. (3x²-2)

Answer 2

Answer :

5 ( x³ - 2x - 1 )⁴ . ( 3x² - 2)

Step-by-step explanation :

In this question, we use the concept of chain rule of differentiation. In order to differentiate complex or composite functions, chain rule is applied.

Chain rule of differentiation is defined as,

• d/dx [f(g(x))] = f'(g(x)) g'(x)

We have to differentiate (x³ - 2x -1 )⁵ with respect to x.

Assume that x³ - 2x - 1 be g(x).

${\small \implies\sf \dfrac{d}{dx} \bigg [ x^3- 2x - 1 \bigg] ^5 = \dfrac{d}{dx} \big[ g(x) \big] . \bigg(\dfrac{ d}{dx} ( x^3) + \dfrac{d}{dx} ( -2x)+ \dfrac{ d}{dx}(-1)\bigg)}$

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We know that,

• $\rm\dfrac{d}{dx} \left( constant\right) = 0$

• $\rm\dfrac{d}{dx} \left( x^{n} \right) = n.x^{(n - 1)}$

• $\rm \frac{ d }{dx} \left( - a^{x} \right) = - (x.a^{x - 1})$

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We get,

${ \implies \small \sf \dfrac{d}{dx} \bigg [ x^3- 2x - 1 \bigg] ^5 = 5 \big[ x^3- 2x - 1 \big] ^4 .( 3 {x}^{2} - 2 + 0)}$

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So the final answer is,

$\rm 5 \big[ x^3- 2x - 1 \big] ^4 .( 3 {x}^{2} - 2 + 0)$