Answer:
Q1:
- For \(f(t)=\dfrac{5}{t-3}-5\): \(f'(t)=-\dfrac{5}{(t-3)^2}\), \(\; f''(t)=\dfrac{10}{(t-3)^3}\).
- For \(f(z)=\ln 100\): \(f'(z)=0\), \(\; f''(z)=0\).
- For \(h(p)=-2e^{\,1-p}\): \(h'(p)=2e^{\,1-p}\), \(\; h''(p)=-2e^{\,1-p}\).
Q1(b) (names and derivatives):
- \(f(t)=\dfrac{5}{t-3}-5\): (reciprocal / rational function) derivatives as above.
- \(f(z)=\ln 100\): (constant function) derivatives 0.
- \(h(p)=-2e^{\,1-p}\): (exponential function) derivatives as above.
Q2:
Assuming the intended function is \(Q(p)=-2\,p^{3/5}\,e^{\,7-p^{7/2}-8p^8}\),
- \(Q'(p)=e^{\,7-p^{7/2}-8p^8}\!\left(-\frac{6}{5}p^{-2/5}+7p^{31/10}+128p^{38/5}\right)\).
- \(Q'(1)=\dfrac{669}{5e^{2}}\).
Explanation:
Q1 Steps:
- For \(5/(t-3)\) write \((t-3)^{-1}\). Derivative: \(5\cdot(-1)(t-3)^{-2}=-5/(t-3)^2\). Second derivative: \(-5\cdot(-2)(t-3)^{-3}=10/(t-3)^3\). Constant \(-5\) differentiates to 0.
- \(\ln 100\) is constant so all derivatives 0.
- For \(h(p)=-2e^{1-p}\): derivative: \(-2\cdot e^{1-p}\cdot(-1)=2e^{1-p}\). Second derivative: \(2\cdot e^{1-p}\cdot(-1)=-2e^{1-p}\).
Q2 Steps:
- Let \(u=-2p^{3/5}\), \(v=e^{7-p^{7/2}-8p^8}\). Then \(u'=-\tfrac{6}{5}p^{-2/5}\). For \(v\), exponent derivative is \(-\tfrac{7}{2}p^{5/2}-64p^7\), so \(v'=v\big(-\tfrac{7}{2}p^{5/2}-64p^7\big)\).
- Use product rule: \(Q'=u'v+uv'\). Factor \(v\) to get the form given.
- Evaluate at \(p=1\): each power equals 1, exponent \(7-1-8=-2\), so bracket is \(-6/5+7+128=\tfrac{669}{5}\). Thus \(Q'(1)=\tfrac{669}{5}e^{-2}=\dfrac{669}{5e^{2}}\).
If your intended form of \(Q(p)\) was different, tell me the exact expression and I will redo the derivative and \(Q'(1)\).