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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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