Question

Hey Can anyone share differentiation formulas of Engineering Maths 1 [MU]

Please Help ! in some days our IA exams are there and i think i haven't studied Much .

Comments

i think it's partial differentiation

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Mujhe Differentiation ke hi chaiye the !

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ohk fir answer dekhlo

Answer 1

In all the formulas below, f’ and g’ represents the following:

• \(\begin{array}{l} \frac{d(f(x))}{dx} = f'(x)\end{array} \)
• \(\begin{array}{l}\frac{d(g(x))}{dx}= g'(x) \end{array} \)

Both f and g are the functions of x and are differentiated with respect to x. We can also represent dy/dx = D_x y. Some of the general differentiation formulas are;

1. Power Rule: (d/dx) (xⁿ ) = nx^n-1
2. Derivative of a constant, a: (d/dx) (a) = 0
3. Derivative of a constant multiplied with function f: (d/dx) (a. f) = af’
4. Sum Rule: (d/dx) (f ± g) = f’ ± g’
5. Product Rule: (d/dx) (fg)= fg’ + gf’
6. Quotient Rule:
   \(\begin{array}{l}\frac{d}{dx}(\frac{f}{g})= \frac{gf’ – fg’}{g^2}\end{array} \)

Differentiation Formulas for Trigonometric Functions

1. \(\begin{array}{l}\frac{d}{dx} (sin~ x)= cos\ x\end{array} \)
2. \(\begin{array}{l}\frac{d}{dx} (cos~ x)= – sin\ x\end{array} \)
3. \(\begin{array}{l}\frac{d}{dx} (tan ~x)= sec^{2} x\end{array} \)
4. \(\begin{array}{l}\frac{d}{dx} (cot~ x = -cosec^{2} x\end{array} \)
5. \(\begin{array}{l}\frac{d}{dx} (sec~ x) = sec\ x\ tan\ x\end{array} \)
6. \(\begin{array}{l}\frac{d}{dx} (cosec ~x)= -cosec\ x\ cot\ x\end{array} \)
7. \(\begin{array}{l}\frac{d}{dx} (sinh~ x)= cosh\ x\end{array} \)
8. \(\begin{array}{l}\frac{d}{dx} (cosh~ x) = sinh\ x\end{array} \)
9. \(\begin{array}{l}\frac{d}{dx} (tanh ~x)= sech^{2} x\end{array} \)
10. \(\begin{array}{l}\frac{d}{dx} (coth~ x)=-cosech^{2} x\end{array} \)
11. \(\begin{array}{l}\frac{d}{dx} (sech~ x)= -sech\ x\ tanh\ x\end{array} \)
12. \(\begin{array}{l}\frac{d}{dx} (cosech~ x ) = -cosech\ x\ coth\ x\end{array} \)

Differentiation Formulas for Inverse Trigonometric Functions

1. \(\begin{array}{l}\frac{d}{dx}(sin^{-1}~ x)=\frac{1}{\sqrt{1 – x^2}}\end{array} \)
2. \(\begin{array}{l}\frac{d}{dx}(cos^{-1}~ x) = -\frac{1}{\sqrt{1 – x^2}}\end{array} \)
3. \(\begin{array}{l}\frac{d}{dx}(tan^{-1}~ x) = \frac{1}{1 + x^2}\end{array} \)
4. \(\begin{array}{l}\frac{d}{dx}(cot^{-1}~ x) = -\frac{1}{1 + x^2}\end{array} \)
5. \(\begin{array}{l}\frac{d}{dx}(sec^{-1} ~x) = \frac{1}{|x|\sqrt{x^2 – 1}}\end{array} \)
6. \(\begin{array}{l}\frac{d}{dx}(cosec^{-1}~x) = -\frac{1}{|x|\sqrt{x^2 – 1}}\end{array} \)

Other Differentiation Formulas

1. \(\begin{array}{l}\frac{d}{dx}(a^{x}) = a^{x} ln a\end{array} \)
2. \(\begin{array}{l}\frac{d}{dx}(e^{x}) = e^{x}\end{array} \)
3. \(\begin{array}{l}\frac{d}{dx}(log_a~ x) = \frac{1}{(ln~ a)x}\end{array} \)
4. \(\begin{array}{l}\frac{d}{dx}(ln~ x) = 1/x\end{array} \)
5. Chain Rule:
   \(\begin{array}{l}\frac{dy}{dx}= \frac{dy}{du}\times \frac{du}{dx}= \frac{dy}{dv}\times \frac{dv}{du}\times \frac{du}{dx}\end{array} \)

Comments

oh ohk Thank you

are ha i think partial ka hi chaiye tha ye nahi ! sorry

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bhai syllabus nahi dekha hai na ?

Ye dekh le Engineering Maths 1 syllabus MU