Question

1. find $\frac{d}{dt}ln 7x^{3}
2. find the equation of the tangent line to y = 5
3. find the equation of the tangent line to g(x)=

an exact value for the slope, do not use a decim

1. evaluate $\frac{d}{dx}\frac{3xln(2x^{3}-5x + 4)}{4 - 3x}$

Explanation:

1. For problem 15: $\frac{d}{dt}\ln(7x^{3})$

• # Explanation:
• ## Step1: Note the variable of differentiation
• We are differentiating with respect to $t$, and $x$ is treated as a constant (assuming $x$ has no relation to $t$). The derivative of a constant - times a function with respect to a variable is the constant times the derivative of the function. The derivative of $\ln(u)$ with respect to $t$ is $\frac{1}{u}\frac{du}{dt}$. Here, $u = 7x^{3}$ and $\frac{du}{dt}=0$ since $7x^{3}$ is a constant with respect to $t$.
• $\frac{d}{dt}\ln(7x^{3})=\frac{1}{7x^{3}}\cdot\frac{d(7x^{3})}{dt}$
• ## Step2: Calculate the derivative
• Since $\frac{d(7x^{3})}{dt}=0$ (as $x$ is a constant with respect to $t$), we have $\frac{d}{dt}\ln(7x^{3}) = 0$.
• # Answer:
• $0$

1. For problem 18: $\frac{d}{dx}\frac{3x\ln(2x^{3}-5x + 4)}{4 - 3x}$

• # Explanation:
• ## Step1: Apply the quotient - rule
• The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 3x\ln(2x^{3}-5x + 4)$ and $v=4 - 3x$.
• First, find $u^\prime$ using the product - rule. The product - rule states that if $u = f(x)g(x)$, then $u^\prime=f^\prime(x)g(x)+f(x)g^\prime(x)$. Let $f(x)=3x$ and $g(x)=\ln(2x^{3}-5x + 4)$.
• $f^\prime(x)=3$ and $g^\prime(x)=\frac{6x^{2}-5}{2x^{3}-5x + 4}$ (by the chain - rule, since the derivative of $\ln(u)$ with respect to $x$ is $\frac{1}{u}\frac{du}{dx}$ and $u = 2x^{3}-5x + 4$).
• So, $u^\prime=3\ln(2x^{3}-5x + 4)+3x\cdot\frac{6x^{2}-5}{2x^{3}-5x + 4}$.
• And $v^\prime=-3$.
• ## Step2: Substitute into the quotient - rule formula
• $\frac{d}{dx}\frac{3x\ln(2x^{3}-5x + 4)}{4 - 3x}=\frac{(3\ln(2x^{3}-5x + 4)+3x\cdot\frac{6x^{2}-5}{2x^{3}-5x + 4})(4 - 3x)-3x\ln(2x^{3}-5x + 4)(-3)}{(4 - 3x)^{2}}$
• Expand the numerator:
• $=\frac{3(4 - 3x)\ln(2x^{3}-5x + 4)+3x(4 - 3x)\frac{6x^{2}-5}{2x^{3}-5x + 4}+9x\ln(2x^{3}-5x + 4)}{(4 - 3x)^{2}}$
• $=\frac{(12 - 9x + 9x)\ln(2x^{3}-5x + 4)+3x(4 - 3x)\frac{6x^{2}-5}{2x^{3}-5x + 4}}{(4 - 3x)^{2}}$
• $=\frac{12\ln(2x^{3}-5x + 4)+3x(4 - 3x)\frac{6x^{2}-5}{2x^{3}-5x + 4}}{(4 - 3x)^{2}}$
• # Answer:
• $\frac{12\ln(2x^{3}-5x + 4)+\frac{3x(4 - 3x)(6x^{2}-5)}{2x^{3}-5x + 4}}{(4 - 3x)^{2}}$

Answer:

1. For problem 15: $\frac{d}{dt}\ln(7x^{3})$

• # Explanation:
• ## Step1: Note the variable of differentiation
• We are differentiating with respect to $t$, and $x$ is treated as a constant (assuming $x$ has no relation to $t$). The derivative of a constant - times a function with respect to a variable is the constant times the derivative of the function. The derivative of $\ln(u)$ with respect to $t$ is $\frac{1}{u}\frac{du}{dt}$. Here, $u = 7x^{3}$ and $\frac{du}{dt}=0$ since $7x^{3}$ is a constant with respect to $t$.
• $\frac{d}{dt}\ln(7x^{3})=\frac{1}{7x^{3}}\cdot\frac{d(7x^{3})}{dt}$
• ## Step2: Calculate the derivative
• Since $\frac{d(7x^{3})}{dt}=0$ (as $x$ is a constant with respect to $t$), we have $\frac{d}{dt}\ln(7x^{3}) = 0$.
• # Answer:
• $0$

1. For problem 18: $\frac{d}{dx}\frac{3x\ln(2x^{3}-5x + 4)}{4 - 3x}$

• # Explanation:
• ## Step1: Apply the quotient - rule
• The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = 3x\ln(2x^{3}-5x + 4)$ and $v=4 - 3x$.
• First, find $u^\prime$ using the product - rule. The product - rule states that if $u = f(x)g(x)$, then $u^\prime=f^\prime(x)g(x)+f(x)g^\prime(x)$. Let $f(x)=3x$ and $g(x)=\ln(2x^{3}-5x + 4)$.
• $f^\prime(x)=3$ and $g^\prime(x)=\frac{6x^{2}-5}{2x^{3}-5x + 4}$ (by the chain - rule, since the derivative of $\ln(u)$ with respect to $x$ is $\frac{1}{u}\frac{du}{dx}$ and $u = 2x^{3}-5x + 4$).
• So, $u^\prime=3\ln(2x^{3}-5x + 4)+3x\cdot\frac{6x^{2}-5}{2x^{3}-5x + 4}$.
• And $v^\prime=-3$.
• ## Step2: Substitute into the quotient - rule formula
• $\frac{d}{dx}\frac{3x\ln(2x^{3}-5x + 4)}{4 - 3x}=\frac{(3\ln(2x^{3}-5x + 4)+3x\cdot\frac{6x^{2}-5}{2x^{3}-5x + 4})(4 - 3x)-3x\ln(2x^{3}-5x + 4)(-3)}{(4 - 3x)^{2}}$
• Expand the numerator:
• $=\frac{3(4 - 3x)\ln(2x^{3}-5x + 4)+3x(4 - 3x)\frac{6x^{2}-5}{2x^{3}-5x + 4}+9x\ln(2x^{3}-5x + 4)}{(4 - 3x)^{2}}$
• $=\frac{(12 - 9x + 9x)\ln(2x^{3}-5x + 4)+3x(4 - 3x)\frac{6x^{2}-5}{2x^{3}-5x + 4}}{(4 - 3x)^{2}}$
• $=\frac{12\ln(2x^{3}-5x + 4)+3x(4 - 3x)\frac{6x^{2}-5}{2x^{3}-5x + 4}}{(4 - 3x)^{2}}$
• # Answer:
• $\frac{12\ln(2x^{3}-5x + 4)+\frac{3x(4 - 3x)(6x^{2}-5)}{2x^{3}-5x + 4}}{(4 - 3x)^{2}}$