Question

The differentiable coefficient of x6 w.r.t x³ is:​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{x^6}$

$\underline{\textsf{To find:}}$

$\textsf{Differential coefficient of}\;\mathsf{x^6}$

$\textsf{with respect to}\;\mathsf{x^3}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{\dfrac{d(x^6)}{d(x^3)}}$

$\mathsf{=\dfrac{\dfrac{d(x^6)}{dx}}{\dfrac{d(x^3)}{dx}}}$

$\mathsf{=\dfrac{6\,x^5}{3\,x^2}}$

$\mathsf{=\dfrac{2\,x^5}{x^2}}$

$\mathsf{=2\,x^3}$

$\underline{\textsf{Answer:}}$

$\mathsf{\dfrac{d(x^6)}{d(x^3)}=2\,x^3}$

Answer 2

SOLUTION :

TO DETERMINE

$\sf{}The \: differentiable \: coefficient \: of \: {x}^{6} \: w.r.t \: \: {x}^{3}$

EVALUATION

Let

$\sf{}y = {x}^{6} \: \: \: \: ....(1)$

$\sf{}z = {x}^{3} \: \: \: \: ....(2)$

Differentiating both sides of Equation (1)with respect x we get

$\displaystyle \sf{} \frac{dy}{dx} = 6 {x}^{5}$

Differentiating both sides of Equation (2) with respect to x we get

$\displaystyle \sf{} \frac{dz}{dx} = 3 {x}^{2}$

$\therefore \sf{}The \: differentiable \: coefficient \: of \: {x}^{6} \: w.r.t \: \: {x}^{3}$

$\displaystyle \sf{} \frac{dy}{dz} = \frac{ \frac{dy}{dx} }{ \frac{dz}{dx} }$

$\displaystyle \sf{} = \frac{6 {x}^{5} }{3 {x}^{2} }$

$\displaystyle \sf{} = 2 {x}^{3}$

FINAL RESULT

$\sf{}The \: differentiable \: coefficient \: of \: {x}^{6} \: w.r.t \: \: {x}^{3} \: is \: 2 {x}^{3}$

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LEARN MORE FROM BRAINLY

Let f(1) = 3,f' (1)=- 1/3,g (1)=-4 and g' (1) =-8/3

The derivative of √[[f(x)]^2+ [g (x)]^2]

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