Answer to Question #150239 in Calculus for Luc

Question

1) Let f(x) = π − √7x − 9. Use the definition of the derivative to find f′(x).
2) Differentiate the following function
g(x)=e^-5x + (1/x −tanx)^3 (√x^2 +1)

Answer 1

1) From the definition of derivative of a function $f(x)$ we know that,

$f'(x)=\lim_{h\to0} [f(x+h)-f(x)]/h$ ....(1)

According to the problem given that,

$f(x)=\pi-\sqrt{(7x-9)}$

So,$f(x+h)=\pi-\sqrt{7(x+h)-9}$

Therefore from (1),we have

$f'(x)=\lim_{h\to0}[f(x+h)-f(x)]/h$ $=lim_{h\to0}[[\pi-\sqrt{7(x+h)-9}]-[\pi-\sqrt{7x-9}]]/h$

$=lim_{h\to0}[\sqrt{7x-9}-\sqrt{7(x+h)-9}]/h$

$=lim_{h\to0}[(\sqrt{7x-9}-\sqrt{7(x+h)-9})×(\sqrt{7x-9}+\sqrt{7(x+h)-9})]\div[h×(\sqrt{7x-9}+\sqrt{7(x+h)-9})]$

$=lim_{h\to0}[(7x-9)-(7(x+h)-9)]\div[h×(\sqrt{7x-9}+\sqrt{7(x+h)-9})]$

$=lim_{h\to0}(-7h)/[h×(\sqrt{7x-9}+\sqrt{7(x+h)-9}]$

$=lim_{h\to0}(-7)/[\sqrt{7x-9}+\sqrt{7(x+h)-9}]$

$=(-7)/[\sqrt{7x-9}+\sqrt{7(x+0)-9}]$

$=(-7)/[2×\sqrt{7x-9}]$

Therefore required $f'(x)=(-7)/[2×\sqrt{7x-9}]$

2)Given that

$g(x)=e^{-5x}+[(1/x)-tanx]^3.\sqrt{x^2+1}$

$g'(x)=d/dx(e^{-5x})+d/dx[[(1/x)-tanx]^3.\sqrt{x^2+1}]$

$=-5.e^{-5x}+[(1/x)-tanx]^3.d/dx[\sqrt{x^2+1}]+\sqrt{x^2+1}.d/dx[(1/x)-tanx]^3$

$=-5.e^{-5x}+[(1/x)-tanx]^3.[1/2{\sqrt{x^2+1}}].d/dx(x^2+1)+\sqrt{x^2+1}×3[(1/x)-tanx]^2.d/dx[(1/x)-tanx]$

$=-5.e^{-5x}+[(1/x)-tanx]^3.[x/\sqrt{x^2+1}]+3\sqrt{x^2+1}×[(1/x)-tanx]^2.[(-1/x^2)-sec^2x]$