Let's solve **problem 13** step-by-step: ### Problem 13 Given: \( f(x) = 2x^9 + 3x^7 - 4x^4 + 7x^2 - x + 1 \) We need to find \( f'(x) \). --- ### **Step-by-Step Solution** This is a polynomial function, so we apply the **power rule**: \[ \frac{d}{dx} [x^n] = n x^{n-1} \] Now, differentiate term by term: 1. \( 2x^9 \) Derivative: \( 2 \cdot 9x^{8} = 18x^8 \) 2. \( 3x^7 \) Derivative: \( 3 \cdot 7x^{6} = 21x^6 \) 3. \( -4x^4 \) Derivative: \( -4 \cdot 4x^{3} = -16x^3 \) 4. \( 7x^2 \) Derivative: \( 7 \cdot 2x^{1} = 14x \) 5. \( -x \) Derivative: \( -1 \) (since \( \frac{d}{dx} x = 1 \)) 6. \( +1 \) Derivative: \( 0 \) (constant term) Add them up: \[ f'(x) = 18x^8 + 21x^6 - 16x^3 + 14x - 1 \] --- ### **Final Answer** \[ \boxed{f'(x) = 18x^8 + 21x^6 - 16x^3 + 14x - 1} \] --- Would you like the solution for the next problem as well? If so, please specify which one!

Certainly! Here's a detailed, step-by-step solution for each problem you listed: --- ### **Problem 13** Given: \[ f(z) = 22^z + 32^z - 4z^* + z - 1 \] *Note:* The notation appears to have typographical errors; assuming the intended functions are exponential functions with bases 2 and 3, and polynomial terms. Based on typical problems, perhaps it is: \[ f(z) = 2^z + 3^z - 4z^* + z - 1 \] But since the exact expressions are unclear, I'll proceed with the most probable intended functions, assuming: \[ f(z) = 2^z + 3^z - 4z + z - 1 \] which simplifies to: \[ f(z) = 2^z + 3^z - 3z - 1 \] **Derivative:** - Derivative of \( 2^z \): \( 2^z \ln 2 \) - Derivative of \( 3^z \): \( 3^z \ln 3 \) - Derivative of \( -3z \): \( -3 \) - Derivative of \( -1 \): \( 0 \) **Answer:** \[ f'(z) = 2^z \ln 2 + 3^z \ln 3 - 3 \] --- ### **Problem 14** Given: \[ y = 8 \sqrt{v} \times t a + 5 \] *Note:* The notation may be off, but assuming: \[ y = 8 v^{1/2} a + 5 \] Assuming variables \( v \) and \( a \) are functions of \( t \), then: \[ \frac{dy}{dt} = 8 \left( \frac{1}{2} v^{-1/2} \frac{dv}{dt} \times a + v^{1/2} \frac{da}{dt} \right) \] If \( v \) and \( a \) are constants, then \( dy/dt = 0 \). Without additional info, derivative expression is: \[ dy/dt = 4 v^{-1/2} a \frac{dv}{dt} + 8 v^{1/2} \frac{da}{dt} \] --- ### **Problem 15** Given: \[ h(z) = t e^z - 2 \] Assuming \( t \) is a constant: \[ h'(z) = t e^z \] --- ### **Problem 16** Given: \[ g(z) = \sqrt{z} \] Derivative: \[ g'(z) = \frac{1}{2} z^{-1/2} \] --- ### **Problem 17** Given: \[ y = a^2 \times 422^{-5} \] Assuming: \[ y = a^2 \times 422^{-5} \] Constant with respect to \( z \). If differentiating with respect to \( a \): \[ dy/da = 2a \times 422^{-5} \] --- ### **Problem 18** Given: \[ z = \sqrt[5]{27 z - 3} \] Rewrite as: \[ z = (27 z - 3)^{1/5} \] Differentiate using chain rule: \[ \frac{dz}{dz} = \frac{1}{5} (27 z - 3)^{-4/5} \times 27 \] Simplify: \[ 1 = \frac{27}{5} (27 z - 3)^{-4/5} \] --- ### **Problem 19** Given: \[ f(z) = 3 e^{z} \] Derivative: \[ f'(z) = 3 e^{z} \] --- ### **Problem 20** Given: \[ h(z) = 2222^z \] Derivative: \[ h'(z) = 2222^z \ln 2222 \] --- ### **Problem 21** Given: \[ y = - \frac{F}{V S \text{ roi}} \] *Note:* Due to unclear notation, assuming it involves differentiating a quotient or product. Please specify. --- ### **Problem 22** Given: \[ f(z) = \ln( 32 z^2 - 22^z + 2 z - T z + 1) \] Derivative: \[ f'(z) = \frac{1}{\text{argument}} \times \text{derivative of argument} \] Calculate numerator: \[ \frac{d}{dz} (32 z^2) = 64 z \] \[ \frac{d}{dz} (-22^z) = -22^z \ln 22 \] \[ \frac{d}{dz} (2 z) = 2 \] \[ \frac{d}{dz} (- T z) = - T \] \[ \frac{d}{dz} (1) = 0 \] Sum: \[ 64 z - 22^z \ln 22 + 2 - T \] Final derivative: \[ f'(z) = \frac{64 z - 22^z \ln 22 + 2 - T}{32 z^2 - 22^z + 2 z - T z + 1} \] --- ### **Problem 23** Given: \[ g(x) = \ln (a^2 x^2 + 7) \] Derivative: \[ g'(x) = \frac{1}{a^2 x^2 + 7} \times 2 a^2 x = \frac{2 a^2 x}{a^2 x^2 + 7} \] --- ### **Problem 24** Given: \[ h(z) = \sin z \cos z \] Use product rule: \[ h'(z) = \cos z \cos z - \sin z \sin z = \cos^2 z - \sin^2 z \] which simplifies to: \[ h'(z) = \cos 2z \] --- ### **Problem 25** Given: \[ y = 2 \csc (22^\circ + 3) + 1 \] Derivative: \[ dy/d\theta = 2 \times (- \csc (22^\circ + 3)) \cot (22^\circ + 3) \times \frac{d}{d\theta} (22^\circ + 3) \] Since 22°+3 is a constant, derivative is zero unless variable is involved. If the variable is in the argument, then: \[ dy/d\theta = - 2 \csc (22^\circ + 3) \cot (22^\circ + 3) \] --- ### **Problem 26** Given: \[ f(t) = 3 t^* - \log (2^t + 3) \] Derivative: \[ f'(t) = 3 \times t^{* -1} - \frac{1}{2^t + 3} \times \frac{d}{dt} (2^t) \] \[ \frac{d}{dt} (2^t) = 2^t \ln 2 \] Thus: \[ f'(t) = 3 t^{* -1} - \frac{2^t \ln 2}{2^t + 3} \] --- ### **Problem 27** Given: \[ g(x) = 3 \arcsin (52^\circ - 3) \] Assuming variable in the argument: \[ g'(x) = 3 \times \frac{1}{\sqrt{1 - (52^\circ - 3)^2}} \times \frac{d}{dx} (52^\circ - 3) \] If the argument is constant, derivative is zero. --- ### **Problem 28** Find the tangent line to: \[ f(z) = 22 z - 3z - 1 \quad \text{at} \quad z=2 \] Simplify: \[ f(z) = 19 z - 1 \] Derivative: \[ f'(z) = 19 \] At \( z=2 \): \[ f(2) = 19 \times 2 - 1 = 38 - 1 = 37 \] Equation of tangent line: \[ y - 37 = 19 (z - 2) \] or: \[ y = 19(z - 2) + 37 = 19z - 38 + 37 = 19z - 1 \] --- ### **Problem 29** Implicit differentiation of: \[ 22 z + 3 y^* = 3 z^* - 22 \] Differentiate both sides with respect to the variable (say \( t \)): \[ 22 \frac{dz}{dt} + 3 y^{* -1} \frac{dy}{dt} = 3 \frac{dz}{dt} \] Rearranged: \[ (22 - 3) \frac{dz}{dt} + 3 y^{* -1} \frac{dy}{dt} = 0 \] Solve for \( dy/dt \): \[ (19) \frac{dz}{dt} + 3 y^{* -1} \frac{dy}{dt} = 0 \] \[ \Rightarrow \frac{dy}{dt} = - \frac{19}{3 y^{* -1}} \frac{dz}{dt} \] --- ### **Problem 30** Given: \[ 30^\circ + 2 u^* v^* + T u^* = 3 \] Differentiate implicitly: \[ 0 + 2 \left( u^* \frac{dv^*}{dt} + v^* \frac{du^*}{dt} \right) + T \frac{du^*}{dt} = 0 \] Solve for \( \frac{du^*}{dt} \): \[ 2 u^* \frac{dv^*}{dt} + 2 v^* \frac{du^*}{dt} + T \frac{du^*}{dt} = 0 \] \[ (2 v^* + T) \frac{du^*}{dt} = - 2 u^* \frac{dv^*}{dt} \] \[ \boxed{ \frac{du^*}{dt} = - \frac{2 u^* \frac{dv^*}{dt}}{2 v^* + T} } \] --- ### **Problems 31 & 32 (Logarithmic differentiation)** **Problem 31:** Given: \[ y = z^c \times [(32\% +1) \tan a] \] Assuming variables involved, the derivative involves: \[ \frac{dy}{dz} = y \times \left( \frac{c}{z} + \text{derivative of the second term} \right) \] --- If you need more specific solutions or clarification on any particular problem, please let me know!

Certainly! Here's a detailed, step-by-step solution for each problem you listed, including calculations and final answers: --- ### **Problem 13** Given: \[ f(z) = 2^z + 3^z - 4z + z - 1 \] **Simplify:** \[ f(z) = 2^z + 3^z - 3z - 1 \] **Differentiate term-by-term:** - \(\frac{d}{dz} 2^z = 2^z \ln 2 \) - \(\frac{d}{dz} 3^z = 3^z \ln 3 \) - \(\frac{d}{dz} (-3z) = -3 \) - \(\frac{d}{dz} (-1) = 0 \) **Final derivative:** \[ \boxed{ f'(z) = 2^z \ln 2 + 3^z \ln 3 - 3 } \] --- ### **Problem 14** Given: \[ y = 8 v^{1/2} a + 5 \] **Assuming \(v\) and \(a\) are functions of \(t\),** then: \[ \frac{dy}{dt} = 8 \left( \frac{1}{2} v^{-1/2} \frac{dv}{dt} \times a + v^{1/2} \frac{da}{dt} \right) \] \[ \Rightarrow \frac{dy}{dt} = 4 v^{-1/2} a \frac{dv}{dt} + 8 v^{1/2} \frac{da}{dt} \] **Final expression:** \[ \boxed{ \frac{dy}{dt} = 4 a v^{-1/2} \frac{dv}{dt} + 8 v^{1/2} \frac{da}{dt} } \] --- ### **Problem 15** Given: \[ h(z) = t e^z - 2 \] **Assuming \(t\) is constant:** \[ h'(z) = t e^z \] **Final answer:** \[ \boxed{ h'(z) = t e^z } \] --- ### **Problem 16** Given: \[ g(z) = \sqrt{z} = z^{1/2} \] **Derivative:** \[ g'(z) = \frac{1}{2} z^{-1/2} = \frac{1}{2 \sqrt{z}} \] **Final answer:** \[ \boxed{ g'(z) = \frac{1}{2 \sqrt{z}} } \] --- ### **Problem 17** Given: \[ y = a^2 \times 422^{-5} \] Assuming \(a\) is variable, differentiate with respect to \(a\): \[ \frac{dy}{da} = 2a \times 422^{-5} \] **Final answer:** \[ \boxed{ \frac{dy}{da} = 2a \times 422^{-5} } \] --- ### **Problem 18** Given: \[ z = \sqrt[5]{27z - 3} = (27z - 3)^{1/5} \] Differentiate using chain rule: \[ \frac{dz}{dz} = \frac{1}{5} (27z - 3)^{-4/5} \times 27 \] Simplify: \[ 1 = \frac{27}{5} (27z - 3)^{-4/5} \] --- ### **Problem 19** Given: \[ f(z) = 3 e^{z} \] Derivative: \[ f'(z) = 3 e^{z} \] --- ### **Problem 20** Given: \[ h(z) = 2222^{z} \] Derivative: \[ h'(z) = 2222^{z} \ln 2222 \] --- ### **Problem 21** Expression appears unclear; unable to process due to notation. --- ### **Problem 22** Given: \[ f(z) = \ln(32 z^2 - 22^z + 2 z - T z + 1) \] Differentiate: \[ f'(z) = \frac{1}{\text{argument}} \times \frac{d}{dz} \left( 32 z^2 - 22^z + 2 z - T z + 1 \right) \] Calculate numerator: - \(\frac{d}{dz} 32 z^2 = 64 z \) - \(\frac{d}{dz} (-22^z) = -22^z \ln 22 \) - \(\frac{d}{dz} 2 z = 2 \) - \(\frac{d}{dz} (-T z) = -T \) - \(\frac{d}{dz} 1 = 0 \) Sum numerator: \[ 64 z - 22^z \ln 22 + 2 - T \] **Final derivative:** \[ \boxed{ f'(z) = \frac{64 z - 22^z \ln 22 + 2 - T}{32 z^2 - 22^z + 2 z - T z + 1} } \] --- ### **Problem 23** Given: \[ g(x) = \ln(a^2 x^2 + 7) \] Derivative: \[ g'(x) = \frac{1}{a^2 x^2 + 7} \times 2 a^2 x = \frac{2 a^2 x}{a^2 x^2 + 7} \] --- ### **Problem 24** Given: \[ h(z) = \sin z \cos z \] Derivative using product rule: \[ h'(z) = \cos z \cos z - \sin z \sin z = \cos^2 z - \sin^2 z \] which simplifies to **cosine double angle:** \[ h'(z) = \cos 2z \] --- ### **Problem 25** Given: \[ y = 2 \csc (22^\circ + 3) + 1 \] If the argument involves a variable, derivative: \[ dy/d\theta = 2 \times (- \csc (22^\circ + 3)) \cot (22^\circ + 3) \times \frac{d}{d\theta} (22^\circ + 3) \] Since \(22^\circ + 3\) is constant, derivative is zero unless variable involved. --- ### **Problem 26** Given: \[ f(t) = 3 t - \log (2^t + 3) \] Derivative: \[ f'(t) = 3 - \frac{1}{2^t + 3} \times 2^t \ln 2 \] Final: \[ \boxed{ f'(t) = 3 - \frac{2^t \ln 2}{2^t + 3} } \] --- ### **Problem 27** Given: \[ g(x) = 3 \arcsin (52^\circ - 3) \] Assuming variable in the argument, derivative: \[ g'(x) = 3 \times \frac{1}{\sqrt{1 - (52^\circ - 3)^2}} \times \frac{d}{dx}(52^\circ - 3) \] If constant, derivative is zero. --- ### **Problem 28** Find the tangent to: \[ f(z) = 22z - 3z - 1 = 19z - 1 \] at \(z=2\): \[ f'(z) = 19 \] \[ f(2) = 19 \times 2 - 1 = 38 - 1 = 37 \] Equation of tangent: \[ y - 37 = 19(z - 2) \Rightarrow y = 19z - 38 + 37 = 19z - 1 \] --- ### **Problem 29** Given: \[ 22 z + 3 y^* = 3 z^* - 22 \] Differentiate implicitly w.r.t. \(t\): \[ 22 \frac{dz}{dt} + 3 y^{* -1} \frac{dy}{dt} = 3 \frac{dz}{dt} \] Rearranged: \[ (22 - 3) \frac{dz}{dt} + 3 y^{* -1} \frac{dy}{dt} = 0 \] \[ 19 \frac{dz}{dt} + 3 y^{* -1} \frac{dy}{dt} = 0 \] Solve for \(dy/dt\): \[ \frac{dy}{dt} = - \frac{19}{3 y^{* -1}} \frac{dz}{dt} = - \frac{19 y}{3} \frac{dz}{dt} \] --- ### **Problem 30** Given: \[ 30^\circ + 2 u^* v^* + T u^* = 3 \] Differentiate: \[ 0 + 2 \left( u^* \frac{dv^*}{dt} + v^* \frac{du^*}{dt} \right) + T \frac{du^*}{dt} = 0 \] Rearranged: \[ (2 v^* + T) \frac{du^*}{dt} = - 2 u^* \frac{dv^*}{dt} \] Final: \[ \boxed{ \frac{du^*}{dt} = - \frac{2 u^* \frac{dv^*}{dt}}{2 v^* + T} } \] --- ### **Problems 31 & 32: Logarithmic Differentiation** **Problem 31:** Given: \[ y = z^c \times [(32\% + 1) \tan a] \] Assuming \(c\) and other variables are functions of \(z\), then: \[ \ln y = c \ln z + \ln \left( (32\% + 1) \tan a \right) \] Differentiating: \[ \frac{1}{y} \frac{dy}{dz} = \frac{dc}{dz} \ln z + c \frac{1}{z} + \text{derivative of the second term} \] --- **Summary:** Due to some notation ambiguities, I provided the most logical interpretations. If you can clarify specific parts, I can refine the solutions further. --- Would you like me to prepare a clean, summarized version of all solutions?