Question

Differentiate w.r.t. d tan(5x+9)/dx

Answer 1

Answer :

Let, y = tan(5x + 9)

Now, differentiating with respect to x, we get

$\frac{dy}{dx} = \frac{d}{dx} tan(5x + 9) \\ \\ \implies \frac{dy}{dx} = {sec}^{2} (5x + 9) \: \frac{d}{dx} (5x + 9) \\ \\ \implies \: \frac{dy}{dx} =5 \: {sec}^{2} (5x + 9)$

RULE :

$\frac{d}{dx}(tanx)=sec^{2}x$

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Answer 2

Answer:

1. Let, y = tan(5x + 9)
3. Now, differentiating with respect to x, we get
5. \begin{gathered} \frac{dy}{dx} = \frac{d}{dx} tan(5x + 9) \\ \\ \implies \frac{dy}{dx} = {sec}^{2} (5x + 9) \: \frac{d}{dx} (5x + 9) \\ \\ \implies \: \frac{dy}{dx} =5 \: {sec}^{2} (5x + 9)\end{gathered}
6. dx
7. dy
9. =
10. dx
11. d
13. tan(5x+9)
14. ⟹
15. dx
16. dy
18. =sec
19. 2
20. (5x+9)
21. dx
22. d
24. (5x+9)
25. ⟹
26. dx
27. dy
29. =5sec
30. 2
31. (5x+9)
35. RULE :
37. \frac{d}{dx}(tanx)=sec^{2}x
38. dx
39. d
41. (tanx)=sec
42. 2
43. x
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