Nine derivatives: chain, product, quotient and inverse functions

Keep π constant and apply the quotient rule. The derivative is $\dfrac\pi{x^2}[x\left(\dfrac12(8100+x^2)^{-1/2}(2x)\right)-(8100+x^2)^{1/2}]$. Factor $(8100+x^2)^{-1/2}$ inside the brackets: the remaining factor is $x^2-(8100+x^2)=-8100$. Thus the derivative is $-\dfrac{8100\pi}{x^2\sqrt{8100+x^2}}$, for $x\ne0$.

$\dfrac d{dx}\sec^3(2x)=3\sec^2(2x)\cdot\sec(2x)\tan(2x)\cdot2=6\sec^3(2x)\tan(2x)$, where $\cos2x\ne0$.

Chain rule gives $\cos\sqrt{x^2+1}\cdot\dfrac12(x^2+1)^{-1/2}(2x)=\dfrac{x\cos\sqrt{x^2+1}}{\sqrt{x^2+1}}$.

Change base: $\log_{10}5x=\ln(5x)/\ln10$. Differentiate: $\dfrac1{\ln10}\cdot\dfrac1{5x}\cdot5=\dfrac1{x\ln10}$, for $x>0$.

The source splits the logarithm first: $\ln[(x^3+2)(x^2+3)]=\ln(x^3+2)+\ln(x^2+3)$. Differentiating gives $\dfrac{3x^2}{x^3+2}+\dfrac{2x}{x^2+3}$. The real logarithm requires $x>-\sqrt[3]2$, since $x^2+3$ is always positive.

The source labels this route “Not Recommended”: differentiate the product inside the logarithm. The derivative is $\dfrac{(x^3+2)(2x)+(x^2+3)(3x^2)}{(x^3+2)(x^2+3)}=\dfrac{5x^4+9x^2+4x}{(x^3+2)(x^2+3)}$. It agrees with the first method on the original domain.

Product rule: $x^2\dfrac d{dx}(e^{x\sin2x})+e^{x\sin2x}\dfrac d{dx}(x^2)$. Differentiate the exponent by the product rule: $(x\sin2x)\prime=2x\cos2x+\sin2x$. Hence the derivative is $x^2(2x\cos2x+\sin2x)e^{x\sin2x}+2xe^{x\sin2x}=xe^{x\sin2x}(2+x\sin2x+2x^2\cos2x)$.

Rewrite as $(\sin^{-1}3x)^{-4}$. The power rule and chain rule give $-4(\sin^{-1}3x)^{-5}\cdot\dfrac3{\sqrt{1-9x^2}}=-\dfrac{12}{(\sin^{-1}3x)^5\sqrt{1-9x^2}}$. This derivative formula applies for $0<|x|<1/3$. The first power-rule line in the printed solution omits the minus sign; its later line and final answer have the correct sign.

$\dfrac d{dx}\sin^{-1}(2\sin x)=\dfrac1{\sqrt{1-(2\sin x)^2}}\cdot2\cos x=\dfrac{2\cos x}{\sqrt{1-4\sin^2x}}$, where $|2\sin x|<1$.

Chain and quotient rules give $\dfrac1{1+((1+x)/(1-x))^2}\cdot\dfrac{(1-x)-(1+x)(-1)}{(1-x)^2}$. This becomes $\dfrac{(1-x)^2}{(1-x)^2+(1+x)^2}\cdot\dfrac2{(1-x)^2}=\dfrac2{1-2x+x^2+1+2x+x^2}=\dfrac1{1+x^2}$. Retain $x\ne1$ from the original input.