Solution: Now on both the sides applying log, we obtain $\log y^{x}=\log x^{y}$ $x \log y=y \log x$ So now, with respect to '$x$' on both sides apply differentiation , we obtain $x\left[\frac{1}{y}...

Solution: Provided $u, v$ and $w$ are the functions of $x$. We need to prove: $\frac{d}{d x}(u \cdot v \cdot w)=\frac{d u}{d x} \cdot v \cdot w+u \cdot \frac{d v}{d x} \cdot w+u \cdot v \cdot...

Solution: Let's take $y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$ (i) Logarithmic differentiation $\begin{aligned} &y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right) \\ &\log y=\log...

Solution: Let's take $y=\left(x^{2}-5 x+8\right)\left(x^{3}+7 x+9\right)$ (i) By using the product rule: $\begin{aligned} &\frac{d y}{d x}=\left(x^{2}-5 x+8\right) \frac{d}{d x}\left(x^{3}+7...

Solution: Provided: $f(x)=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)\left(1+x^{8}\right)$...........(1) $\log f(x)=\log (1+x)+\log \left(1+x^{2}\right)+\log \left(1+x^{4}\right)+\log...

Solution: Provided: $x y=e^{e-x}$ $\log x y=\log e^{x-y}$ $\log x+\log y=(x-y) \log e$ $\log x+\log y=(x-y)[\because \log e=1]$ $\frac{d}{d x} \log x+\frac{d}{d x} \log y=\frac{d}{d x}(x-y)$...

Solution: Provided: $(\cos x)^{y}=(\cos y)^{x}$ $\log (\cos x)^{y}=\log (\cos y)^{x}$ $y \log \cos x=x \log \cos y$ $\frac{d}{d x}(y \log \cos x)=\frac{d}{d x}(x \log \cos y)$ $y \frac{d}{d x} \log...

Solution: Provided: $x^{y^{y}}+y^{x}=1$ $u+v=1 \text {, in which } u=x^{y} \text { and } v=y^{z}$ $\frac{d}{d x} u+\frac{d}{d x} v=\frac{d}{d x} 1$ $\frac{d u}{d x}+\frac{d v}{d x}=0 \dots \dots...

Solution: Let's take $y=(x \cos x)^{x}+(x \sin x)^{\frac{1}{2}}$ Putting the value of $u=(x \cos x)^{x}$ and ${ }^{v=(x \sin x)^{\frac{1}{x}}}$, we obtain $y=u+v$ $\frac{d y}{d x}=\frac{d u}{d...

Solution: Let's take $x^{x^{x \cos x}+\frac{x^{2}+1}{x^{2}-1}}$ Putting the value of $u=x^{x \operatorname{cox} x}$ and $v=\frac{x^{2}+1}{x^{2}-1}$, We obtain $y=u+v$ $\frac{d y}{d x}=\frac{d u}{d...

Solution: Let's take $y=x^{\sin x}+(\sin x)^{\cos x}$ Putting the value of $u=x^{\sin x} \text { and } v=(\sin x)^{\cos x}$, We obtain $y=u+\nu$ $\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}...

Solution: Let's take $y=(\sin x)^{x}+\sin ^{-1} \sqrt{x}=u+v$ In which $u=(\sin x)^{x} \text { and } v=\sin ^{-1} \sqrt{x}$ $\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$.........(1) So now...

Solution: Let's take $y=(\log x)^{x}+x^{\log x}=u+v$ In which $u=(\log x)^{x}$ and $v=x^{\log x}$ $\frac{d y}{d x}=\frac{d u}{d x}+\frac{d v}{d x}$ So now $u=(\log x)^{x}$ $\begin{array}{l} \log...

Solution: Let's take $^{y}=\left(x+\frac{1}{x}\right)^{x}+x^{\left(x-\frac{1}{x}\right)}$ Putting the value of $\left(x+\frac{1}{x}\right)^{x}=u$ and $x^{\left(x+\frac{1}{x}\right)}=v$ $y=u+v$...

Solution: Let's take $y=(x+3)^{2}(x+4)^{3}(x+5)^{4}$ On both the sides taking log, we get $\log y=2 \log (x+3)+3 \log (x+4)+4 \log (x+5)^{4}$ So now, $\begin{aligned} &\frac{d}{d x} \log y=2...

Solution: Let's take $y=x^{x}-2^{\sin x}$ Put $u=x^{x} \text { and } v=2^{\text {sinx }}$ $y=u-v$ $\frac{d y}{d x}=\frac{d u}{d x}-\frac{d v}{d x} \ldots \dots \dots$(1) Now, $u=x^{2}$ $\log u=\log...

Solution: Let's take $y=(\log x)^{\cos x}$ On both the sides taking log, we get $\log y=\log (\log x)^{\cos x}=\cos x \log (\log x)$ $\frac{d}{d x} \log y=\frac{d}{d x}[\cos x \log (\log x)]$...

Solution: Let's take $y=\sqrt{\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}}$ $=\left(\frac{(x-1)(x-2)}{(x-3)(x-4)(x-5)}\right)^{\frac{1}{2}}$ On both the sides taking log, we get $\log y=\frac{1}{2}[\log...

Solution: Let's take $y=\cos x \cos 2 x \cos 3 x$ On both the sides taking log, we get $\log y=\log (\cos x \cos 2 x \cos 3 x)$ $=\log \cos x+\log \cos 2 x+\log \cos 3 x$ So now, $\frac{d}{d x} \log...