Question

diff.y=(3x^3+5x^2+8)^3​

Answer 1

Question :

To Find Differention of

$y = (3x {}^{3} + 5x {}^{2} + 8) {}^{3}$

Theory :

•Chain Rule :

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

• Differentiation Formula's :

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

$2) \dfrac{d(constant)}{dx} = 0$

Solution :

we have to find $\dfrac{dy}{dx}$

$y = (3x {}^{3} + 5x {}^{2} + 8) {}^{3}$

Let$u=(3x {}^{3}+ 5x {}^{2}+8)$

$\implies \: y = (u) {}^{3}$

Now Differentiate with respect to x

$\implies \dfrac{dy}{dx} = \dfrac{d(u) {}^{3} }{du} \times \dfrac{du}{dx}$

$\implies \dfrac{dy}{dx} = 3u {}^{2} \times \dfrac{du}{dx}$

$\implies \dfrac{dy}{dx} = 3(3x {}^{3} + 5x {}^{2} + 8){}^{2} \times (9x {}^{2} + 10x)$

which is the required solution!