Calculas_img

(i)(3x-5)\displaystyle ^{4}
মনে করি,
y= (3x-5)\displaystyle ^{4}
X এর সাপেক্ষে অন্তরীকরণ করে পাই ,
\displaystyle \frac{dy}{dx}=\displaystyle \frac{d}{dx}(3x-5)\displaystyle ^{4}
=4(3x-5)\displaystyle ^{4-1}.\displaystyle \frac{d}{dx}(3x-5)
=4(3x-5)\displaystyle ^{3}.\displaystyle \frac{d}{dx}(3x)-\displaystyle \frac{d}{dx}(5)
=4(3x-5)\displaystyle ^{3}.(3-0)
=4(3x-5)\displaystyle ^{3}.3
=12(3x-5)\displaystyle ^{3} (ans:)

Calculas_img

(ii)e\displaystyle ^{3x}
মনে করি ,
y=e\displaystyle ^{3x}
X এর সাপেক্ষে অন্তরীকরণ করে পাই ,
\displaystyle \frac{dy}{dx}=\displaystyle \frac{d}{dx}(e\displaystyle ^{3x})
= e\displaystyle ^{3x}.\displaystyle \frac{d}{dx} (3x)
= e\displaystyle ^{3x}.3
=3e\displaystyle ^{3x} (ans:)

Calculas_img

(iii)e\displaystyle ^{\sqrt{x}}
মনে করি,
y=e\displaystyle ^{\sqrt{x}}
X এর সাপেক্ষে অন্তরীকরণ করে পাই,
\displaystyle \frac{dy}{dx} =\frac{d}{dx}(e\displaystyle ^{\sqrt{x}})
=e\displaystyle ^{\sqrt{x}}.\displaystyle \frac{d}{dx}(\displaystyle \sqrt{x})
=e\displaystyle ^{\sqrt{x}}.\displaystyle \frac{1}{2\sqrt{x}}
=\displaystyle \frac{e^{\sqrt{x}}}{2\sqrt{x}} (ans:)

Calculas_img

(iv)e\displaystyle ^{sinx}
মনে করি,
y=e\displaystyle ^{sinx}
X এর সাপেক্ষে অন্তরীকরণ করে পাই,
\displaystyle \frac{dy}{dx} =\frac{d}{dx}( e^{sinx})
=e\displaystyle  \begin{array}{{>{\displaystyle}l}} ^{sinx}\frac{d}{dx}( sinx)\ \end{array}
=e\displaystyle ^{sinx}.cosx (ans:)

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(v) sin(ax)
মনে করি,
y= sin(ax)
X এর সাপেক্ষে অন্তরীকরণ করে পাই,
\displaystyle \frac{dy}{dx}=\displaystyle \frac{d}{dx}sin(ax)
=cos(ax)\displaystyle \frac{d}{dx}(ax)
= cos(ax)..a
=acos(ax) (ans):

Calculas_img


(vi) sec(5x+3)
মনে করি,
y=sec(5x+3)
X এর সাপেক্ষে অন্তরীকরণ করে পাই,
\displaystyle \frac{dy}{dx}=\displaystyle \frac{d}{dx}sec(5x+3)
=sec(5x+3).tan(5x+3)\displaystyle \frac{d}{dx}(5x+3)
=sec(5x+3).tan(5x+3)\displaystyle \frac{d}{dx}(5x)+\displaystyle \frac{d}{dx}(3)
=sec(5x+3).tan(5x+3).(5-0)
= 5 sec(5x+3).tan(5x+3) (ans):

Calculas_img

(vii) tan(ax+b)
মনে করি,
y=tan(ax+b)
X এর সাপেক্ষে অন্তরীকরণ করে পাই,
\displaystyle  \begin{array}{{>{\displaystyle}l}} \frac{dy}{dx} =\frac{d}{dx} tan( ax+b)\ \ \ \ \ =sec^{2}( ax+b)\frac{d}{dx}( ax+b)\ \ \ \ \ \ =sec^{2}( ax+b)\frac{d}{dx}( ax) +\frac{d}{dx}( b)\ \ \ \ \ \ \ =sec^{2}( ax+b) .( a+0)\ \ \ \ \ \ \ \ =a\ sec^{2}( ax+b) \ \ ( ans) : \end{array}

calculas_img

(vii) cosx\displaystyle ^{0}
মনে করি,
y=cosx\displaystyle ^{0}
X এর সাপেক্ষে অন্তরীকরণ করে পাই,
\displaystyle \frac{dy}{dx}=\displaystyle \frac{d}{dx}(cosx\displaystyle ^{0})
=\displaystyle \frac{d}{dx}(cox\displaystyle \frac{\prod x}{180^{0}})
=-(sin\displaystyle \frac{\prod x}{180^{0}})\displaystyle \frac{d}{dx}\left(\frac{\prod x}{180^{0}}\right)
=-(sin\displaystyle \frac{\prod x}{180^{0}}).\displaystyle \frac{\prod }{180^{0}}
= –\displaystyle \frac{\prod }{180^{0}}sin\displaystyle \frac{\prod x}{180^{0}} (ans):

Calculas_img

viii) x\displaystyle \sqrt{sinx}
মনে করি,
y= x\displaystyle \sqrt{sinx}
X এর সাপেক্ষে অন্তরীকরণ করে পাই,
\displaystyle \frac{dy}{dx}=\displaystyle \frac{d}{dx}(x\displaystyle \sqrt{sinx})
=x\displaystyle \frac{d}{dx} \displaystyle (\sqrt{sinx})+(\displaystyle \sqrt{sinx)}\displaystyle \frac{d}{dx}(x)
(বাকি অংশ)


Post Author: showrob

1 thought on “

    soikot

    (March 28, 2020 - 3:27 pm)

    very nice

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