Solution. Each is the chain rule applied to a standard derivative.

(a). With \(u=x^{2}\), \(\frac {du}{dx}=2x\), so \[\frac {dy}{dx}=\frac {-1}{\sqrt {1-u^{2}}}\cdot 2x =\frac {-2x}{\sqrt {1-x^{4}}} .\]

(b). With \(u=e^{3x}\), \(\frac {du}{dx}=3e^{3x}\), so \[\frac {dy}{dx}=\frac {3e^{3x}}{1+\left (e^{3x}\right )^{2}} =\frac {3e^{3x}}{1+e^{6x}} .\]

(c). With \(u=\arcsin x\), \(y=u^{2}\), so \[\frac {dy}{dx}=2u\cdot \frac {1}{\sqrt {1-x^{2}}} =\frac {2\arcsin x}{\sqrt {1-x^{2}}} .\]

(d). With \(u=(x+2)^{-1}\), \(\frac {du}{dx}=-(x+2)^{-2}\), so \[\frac {dy}{dx}=\frac {-1}{\sqrt {1-\frac {1}{(x+2)^{2}}}}\cdot \left (\frac {-1}{(x+2)^{2}}\right ) =\frac {1}{(x+2)^{2}}\cdot \frac {\left |x+2\right |}{\sqrt {(x+2)^{2}-1}} =\frac {1}{\left |x+2\right |\sqrt {(x+2)^{2}-1}} .\] The absolute value appears because \(\sqrt {(x+2)^{2}}=\left |x+2\right |\), not \(x+2\) — the expression is defined for \(x+2\leq -1\) as well as \(x+2\geq 1\).

Solution. (a). \(\frac {dy}{dx}=\frac {3x^{2}}{1+x^{6}}\).

(b). \(\frac {dy}{dx}=2\ln x\cdot \frac {1}{x}=\frac {2\ln x}{x}\).

(c). Write \(y=(4x+3)(2x-1)^{-1/2}\) and use the product rule: \begin {align*} \frac {dy}{dx} &=4(2x-1)^{-1/2}+(4x+3)\left (-\tfrac {1}{2}\right )(2x-1)^{-3/2}\cdot 2\\ &=(2x-1)^{-3/2}\left [4(2x-1)-(4x+3)\right ]\\ &=\frac {8x-4-4x-3}{(2x-1)^{3/2}} =\frac {4x-7}{(2x-1)^{3/2}} . \end {align*}

Taking the common factor \((2x-1)^{-3/2}\) out before simplifying is what keeps this short; expanding first produces an unnecessary mess.

Solution. (a). With \(y=x\left (1+x^{4}\right )^{-1/2}\), \begin {align*} \frac {dy}{dx} &=\left (1+x^{4}\right )^{-1/2} +x\left (-\tfrac {1}{2}\right )\left (1+x^{4}\right )^{-3/2}\cdot 4x^{3}\\ &=\left (1+x^{4}\right )^{-3/2}\left [\left (1+x^{4}\right )-2x^{4}\right ] =\frac {1-x^{4}}{\left (1+x^{4}\right )^{3/2}} . \end {align*}

(b). Differentiate implicitly, remembering that \(y\) is a function of \(x\): \[\sec ^{2}(x+y)\left (1+\frac {dy}{dx}\right )=1 \quad \Longrightarrow \quad 1+\frac {dy}{dx}=\frac {1}{\sec ^{2}(x+y)}=\frac {1}{1+\tan ^{2}(x+y)} .\] Since \(\tan (x+y)=x\), this is \(\frac {1}{1+x^{2}}\), so \[\frac {dy}{dx}=\frac {1}{1+x^{2}}-1=\frac {-x^{2}}{1+x^{2}} .\]

(c). A variable raised to a variable power: take logarithms first. \[\ln y=\cos \left (x^{2}\right )\ln x .\] Differentiating both sides, \[\frac {1}{y}\frac {dy}{dx} =-2x\sin \left (x^{2}\right )\ln x+\frac {\cos \left (x^{2}\right )}{x},\] so \[\frac {dy}{dx}=x^{\cos x^{2}} \left [\frac {\cos \left (x^{2}\right )}{x}-2x\sin \left (x^{2}\right )\ln x\right ].\] Neither the power rule nor the exponential rule applies directly here, because the base and the exponent both vary. Logarithmic differentiation is the standard escape.

(d). Convert the base: \(\log _{10}(x+1)=\frac {\ln (x+1)}{\ln 10}\), so \[\frac {dy}{dx}=\frac {1}{\ln 10}\left [\ln (x+1)+\frac {x}{x+1}\right ].\]

Solution. (a). With \(y=\left (x^{2}-2\right )(x+2)^{-1/2}\), \begin {align*} \frac {dy}{dx}&=2x(x+2)^{-1/2}-\tfrac {1}{2}\left (x^{2}-2\right )(x+2)^{-3/2}\\ &=\frac {(x+2)^{-3/2}}{2}\left [4x(x+2)-\left (x^{2}-2\right )\right ] =\frac {3x^{2}+8x+2}{2(x+2)^{3/2}} . \end {align*}

(b). Logarithmic differentiation again, since the function is a product and quotient of awkward pieces: \[\ln y=x\cos \left (x^{2}+2\right )-2\ln (x-2).\] Differentiating, and noting \(\frac {d}{dx}\left [x\cos \left (x^{2}+2\right )\right ] =\cos \left (x^{2}+2\right )-2x^{2}\sin \left (x^{2}+2\right )\), \[\frac {1}{y}\frac {dy}{dx} =\cos \left (x^{2}+2\right )-2x^{2}\sin \left (x^{2}+2\right )-\frac {2}{x-2},\] so \[\frac {dy}{dx}=\frac {e^{x\cos \left (x^{2}+2\right )}}{(x-2)^{2}} \left [\cos \left (x^{2}+2\right )-2x^{2}\sin \left (x^{2}+2\right )-\frac {2}{x-2}\right ].\]

Solution. (a). By the product rule, \[\frac {dy}{dx}=e^{3x}+3xe^{3x}=e^{3x}(1+3x).\]

(b). The base is constant and the exponent varies, so \(\ln y=x^{3}\ln 2\) and \[\frac {1}{y}\frac {dy}{dx}=3x^{2}\ln 2 \quad \Longrightarrow \quad \frac {dy}{dx}=3x^{2}\ln 2\cdot 2^{x^{3}} .\] Compare with part (c) of the previous problem: there both base and exponent varied and two terms appeared. Here only one does. The \(\ln 2\) is easy to drop and is what makes \(\frac {d}{dx}2^{x}=2^{x}\ln 2\) rather than \(2^{x}\).

Solution. Let \(\theta =\arcsin \frac {\sqrt {3}}{2}\), so \(\sin \theta =\frac {\sqrt {3}}{2}\) with \(\theta \) in \(\left [-\frac {\pi }{2},\frac {\pi }{2}\right ]\), giving \(\theta =\frac {\pi }{3}\).

The equation asks for \(\cos \theta \), not \(\sin \theta \): \[\cos \frac {\pi }{3}=\frac {1}{2},\] so \(\frac {1}{2}=x+\sqrt {2}\) and \[x=\frac {1}{2}-\sqrt {2}\approx -0.914 .\]