Question

Find the derivative of the following

(A) Y=4t^3 - 5t^2 + 6t-

(B) F(x)= 1/x4 + \sqrt{x}

Answer 1

Given :

A. y = 4t³ - 5t² + 6t

B. f(x) = 1/x⁴ + √x

To Find :

Derivative of the given

Solution :

A.

y = 4t³ - 5t² + 6t

Differentiate with respect to t on both sides ,

➠  $\sf \dfrac{dy}{dt}=\dfrac{d}{dt}(4t^3-5t^2+6t)$

$\bullet\ \; \sf \dfrac{d}{dx}\big(x^n\big)nx^{n-1}$

➠   $\sf \dfrac{dy}{dt}$  = 4(3t²) - 5(2t) + 6

➠   $\sf \dfrac{dt}{dx}$  = 12t² - 10t + 6

➠  $\sf \dfrac{dy}{dt}$   = 2 ( 6t² - 5t + 3 )

So ,

$\sf \dfrac{dy}{dt}=2(6t^2-5t+3)\ \; \orange{\bigstar}$

B.

$\sf f(x)=\dfrac{1}{x^4}+\sqrt{x}$

$\to \sf f(x)=x^{-4}+\sqrt{x}$

Differentiate with respect to x on both sides ,

$\to \sf f'(x)=-4(x^{-4-1})+\dfrac{1}{2\sqrt{x}}\\\\\to \sf f'(x)=-\dfrac{4}{x^5}+\dfrac{1}{2\sqrt{x}}\ \; \green{\bigstar}$

Answer 2

Solution:

• 1st

$y = 4 {t}^{3} - 5 {t}^{2} + 6t$

$\dfrac{dy}{dt} = \dfrac{d \: 4 {t}^{3} - 5 {t}^{2} + 6t }{dt}$

$\dfrac{dy}{dt} = (3 \times 4 {t}^{2} ) - (5 \times 2 \times t) + (6 \times 1 {t}^{0} )$

$\red{\bold{\boxed{\large{\dfrac{dy}{dt} = 12 {t}^{2} - 10t + 6}}}}$

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• 2nd

f(x)= $\dfrac{1}{ {x}^{4} } + \sqrt{x}$

or f(x) $= {x}^{ - 4} + {x}^{ \frac{1}{2} }$

Differentiating we get:

$\dfrac{ {x}^{ - 4} }{dx} + {x}^{ \dfrac{1}{2} }$

$= - 4 {x}^{ - 4 - 1} + \dfrac{1}{2} {x}^{ \frac{1}{2} - 1 }$

$= - 4 {x}^{ - 5} + \dfrac{1}{2} {x}^{ - \frac{1}{2} }$

$\red{\boxed{\bold{\large{= \dfrac{ - 4}{ {x}^{5} } + \dfrac{1}{2 \sqrt{x} }}}}}$

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Formula used:

• $\dfrac{ {a}^{b}}{da}= {a}^{ b-1}$