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Monday, 20 November 2017

Differentiation Formulas

Differentiation Formulas


In the following u,v represent functions of x while a,c,p represent constants . We assume of course that the derivatives of u and v exist, i.e. u and v are differentiable .

1. \cfrac{\mathrm{d}}{\mathrm{d}x}(u\pm v)=\cfrac{\mathrm{d}u}{\mathrm{d}x}\pm \cfrac{\mathrm{d}v}{\mathrm{d}x}

2. \cfrac{\mathrm{d}}{\mathrm{d}x}(cu)=c\cfrac{\mathrm{d}u}{\mathrm{d}x}

3. \cfrac{\mathrm{d}}{\mathrm{d}x}(uv)=u\cfrac{\mathrm{d}v}{\mathrm{d}x}+v\cfrac{\mathrm{d}u}{\mathrm{d}x}

4. \cfrac{\mathrm{d}}{\mathrm{d}x}\left(\cfrac{u}{v}\right) = \cfrac{v(\mathrm{d}u)/\mathrm{d}x -u(\mathrm{d}v/\mathrm{d}x)}{v^2}

5. \cfrac{\mathrm{d}}{\mathrm{d}x}u^p =pu^{p-1}\cfrac{\mathrm{d}u}{\mathrm{d}x}

6. \cfrac{\mathrm{d}}{\mathrm{d}x}=a^u\ln a

7. \cfrac{\mathrm{d}}{\mathrm{d}x}e^u = e^u\cfrac{\mathrm{d}u}{\mathrm{d}x}

8. \cfrac{\mathrm{d}}{\mathrm{d}x}\ln u=\cfrac{1}{u}\cfrac{\mathrm{d}u}{\mathrm{d}x}

9. \cfrac{\mathrm{d}}{\mathrm{d}x}\sin u=\cos u\cfrac{\mathrm{d}u}{\mathrm{d}x}

10. \cfrac{\mathrm{d}}{\mathrm{d}x}\cos u=-\sin u\cfrac{\mathrm{d}u}{\mathrm{d}x}

11. \cfrac{\mathrm{d}}{\mathrm{d}x}\tan u=\sec ^2u\cfrac{\mathrm{d}u}{\mathrm{d}x}

12. \cfrac{\mathrm{d}}{\mathrm{d}x}\cot u=-\csc ^2u\cfrac{\mathrm{d}u}{\mathrm{d}x}

13. \cfrac{\mathrm{d}}{\mathrm{d}x}\sec u=\sec u\tan u\cfrac{\mathrm{d}u}{\mathrm{d}x}

14. \cfrac{\mathrm{d}}{\mathrm{d}x}\csc u=-\csc u \cot u\cfrac{\mathrm{d}u}{\mathrm{d}x}

15. \cfrac{\mathrm{d}}{\mathrm{d}x}\sin^{-1}u= \cfrac{1}{\sqrt{1-u^3}}\cfrac{\mathrm{d}u}{\mathrm{d}x}

16. \cfrac{\mathrm{d}}{\mathrm{d}x}\cos ^{-1}u=\cfrac{-1}{\sqrt{1-u^2}}\cfrac{\mathrm{d}u}{\mathrm{d}x}

17. \cfrac{\mathrm{d}}{\mathrm{d}x}\tan^{-1} u=\cfrac{1}{1+u^2}\cfrac{\mathrm{d}u}{\mathrm{d}x}

18. \cfrac{d}{\mathrm{d}x}\cot ^{-1} u=\cfrac{-1}{1+u^2}\cfrac{\mathrm{d}u}{\mathrm{d}x}

19. \cfrac{\mathrm{d}}{\mathrm{d}x}\mathrm{sinh} \mathrm{cosh} u\cfrac{\mathrm{d}u}{\mathrm{d}x}

20 \cfrac{\mathrm{d}}{\mathrm{d}x} \mathrm{cosh} u=\mathrm{sinh} u\cfrac{\mathrm{d}u}{\mathrm{d}x}

In the special case where u=x, the above formulas are simplified since in such case \cfrac{\mathrm{d}u}{\mathrm{d}x}=1.

by. theory and problems of advanced mathematics

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