Question

Find the nth differential coefficients of sin(x)log(ax+b)

Answer 1

Answer:

sin(x)log(ax+b)

Step-by-step explanation:

hope you happy bro..

Answer 2

Answer:

The nth differential coefficient of $sin(x)$ is given by $cos x$

The nth differential coefficient of $log(ax+b)$ is given by $\frac{d^{n} y}{d x^{2}} =\frac{(-1)^{n-1} \cdot(n-1) ! a^{n}}{(a x+b)^{n}}$

Step-by-step explanation:

a) Solution for Sin x

$f(x)=\sin x \\\therefore f(x+h)=\sin (x+h) \\\therefore f(x+h)-f(x)=sin (x+h)-\sin x \Rightarrow \frac{f(x+h)-f(x)}{h}\\=\frac{\sin (x+h)-\sin x}{h} \Rightarrow \lim _{h \rightarrow 0} \\\frac{f(x+h)-f(x)}{h}=\lim _{h \rightarrow 0}\\ \frac{\sin (x+h)-\sin x}{h}$

$\Rightarrow \frac{d}{d x} f(x)=\lim _{h \rightarrow 0}\left[\frac{2 \cos \left(x+\frac{h}{2}\right) \cdot \sin \frac{h}{2}}{h}\right] \\ \Rightarrow \frac{d}{d x} \sin x=\lim _{h \rightarrow 0} \cos \left(x+\frac{h}{2}\right) \cdot \lim _{h \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right)\\ =\cos (x+0) \cdot \lim _{\frac{h}{2} \rightarrow 0}\left(\frac{\sin \frac{h}{2}}{\frac{h}{2}}\right) \cdots h \rightarrow 0$

$\therefore \frac{h}{2} \rightarrow 0 \\ =\cos x .1\\ \Rightarrow \frac{d}{d x} \sin x=\cos x$

b) Let $y=\log (a x+b)$

Then $\frac{d y}{d x}=\frac{d}{d x}[\log (a x+b)]$

$=\frac{1}{a x+b} \cdot \frac{d}{d x}(a x+b)\\ =\frac{1}{a x+b} \times(a \times 1+0)\\ =\frac{a}{a x+b} \frac{d^{2} y}{d x^{2}}\\=\frac{d}{d x}\left(\frac{a}{a x+b}\right)\\ =a \frac{d}{d x}(a x+b)^{-1}$

$\frac{d^{2} y}{d x^{2}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\frac{a}{a x+b}\right)\\ =a \frac{\mathrm{d}}{\mathrm{dx}}(a x+b)^{-1}\\ =a(-1)(a x+b)^{-2} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(a x+b)\\ =\frac{(-1) a}{(a x+b)^{2}} \times(a \times 1+0)\\ =\frac{(-1) a}{(a x+b)^{2}}$

$\frac{d^{3} y}{d x^{3}}=\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{(-1)^{1} a^{2}}{(a x+b)^{2}}\right]\\ =(-1)^{1} a^{2} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(a x+b)^{-2}\\ =(-1)^{1} a^{2} \cdot(-2)(a x+b)^{-3} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(a x+b)\\ =\frac{(-1)^{2} \cdot 1 \cdot 2 \cdot a^{2}}{(a x+b)^{3}} \times(a \times 1+0)\\ =\frac{(-1) \cdot 2 \cdot 2 ! a^{3}}{(a x+b)^{3}}$

In general, the $\mathrm{n}^{\text {th }}$ order derivative is given by

$\frac{d^{n} y}{d x^{2}}=\frac{(-1)^{n-1} \cdot(n-1) ! a^{n}}{(a x+b)^{n}}$