Question

y=√x(x^2+7) find Dy/dx
​

Answer 1

Explanation:

Differentiating w.r.t x,

dy/dx=√x.d/dx(x^2+7)+(x^2+7).d/dx(√x)

dy/dx=√x.2x+1/2√x.x^2+1/2√x.7

Answer 2

$\huge{\underline{\bf\red{Questi {\mathbb{O}} n} :}}$

• Find $\sf \dfrac{dy}{dx}$

$\huge{\underline{\: \bf\green{Answ {\mathbb{E}} r}\ \: :}}$

$\boxed{\frac{5x { \big( \sqrt{x} \big)}^{3} + 7 \sqrt{x} }{2x} }$

$\Large{\underline{\bf\purple{Giv {\mathbb{E}} n}\ :}}$

• $\sf y = \sqrt{x} \big( {x}^{2} + 7 \big)$

$\huge{\underline{\bf\blue{Soluti {\mathbb{O}} n}\ :}}$

$\begin{aligned} & \quad \ \sf \frac{dy}{dx} \\ \\ & = \sf \frac{d}{dx} \bigg\{ \sqrt{x} \big( {x}^{2} + 7\big) \bigg\} \\ \\ & = \sqrt{x} \frac{ \sf d}{{ \sf{d}}x}( {x}^{2} + 7) + ( {x}^{2} + 7) \frac{ \sf d}{{ \sf{d}}x} ( \sqrt{x} ) \\ \\ & = \sqrt{x} \{ 2 {x}^{2 - 1} + 0 \} + ( {x}^{2} + 7) \bigg( \frac{1}{2} {x}^{ \frac{1}{2} - 1} \bigg) \\ \\ & = {x}^{ \frac{1}{2} }(2x) + ( {x}^{2} + 7) \bigg( \frac{1}{2} {x}^{ - \frac{1}{2} } \bigg) \\ \\ & = 2 {x}^{ \frac{3}{2} } + \frac{1}{2} {x}^{ \frac{3}{2} } + \frac{7}{2} {x}^{ - \frac{1}{2} } \\ \\ & = 2 {( \sqrt{x} )}^{3} + \frac{1}{2 } {( \sqrt{x}) }^{3} + \frac{7}{2 \sqrt{x} } \\ \\ & = \frac{5}{2} {( \sqrt{x} )}^{3} + \frac{7 \sqrt{x} }{2x} \\ \\ & = \frac{5x { \big( \sqrt{x} \big)}^{3} + 7 \sqrt{x} }{2x} & \end{aligned}$

$\red{\rule{5.5cm}{0.02cm}}$

Differentiation formulae Used :--

$\boxed{ \frac{ \sf d}{{\sf{d}}x} {(x)}^{n} = n {x}^{(n - 1)} }$

$\boxed{ \frac{ \sf d}{{ \sf{d}}x}(constant) = 0 }$

$\boxed{ \frac{ \sf d}{{ \sf{d}}x}(u + v) = \frac{ \sf d}{{ \sf{d}}x}(u) + \frac{ \sf d}{{ \sf{d}}x}(v) }$

$\boxed{\frac{ \sf d}{{ \sf{d}}x}(uv) = u\frac{ \sf d}{{ \sf{d}}x}(v) + v\frac{ \sf d}{{ \sf{d}}x}(u)}$

$\purple{\rule{7.5cm}{0.02cm}}$

$\Large{ \mid {\underline{\underline{\bf\green{BrainLiest \ AnswEr}}}} \mid }$