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Find the derivative of the function \(g(t)\), below. It may be to your advantage to simplify before differentiating. \(g(t)=\cos (\ln (t))\) \(g^{\prime}(t)=\) _________

Short Answer

Expert verified
The derivative is \( g'(t) = -\frac{\text{sin}(\text{ln}(t))}{t} \).

Step by step solution

01

Identify the Outer and Inner Functions

The function given is a composition of two functions: the outer function is \(\text{cos}(u)\) and the inner function is \(\text{ln}(t)\).
02

Apply the Chain Rule

To differentiate \(g(t) = \text{cos}(\text{ln}(t))\), use the chain rule: \[ g'(t) = \frac{d}{dt}[\text{cos}(\text{ln}(t))] = \frac{d}{du}[\text{cos}(u)] \times \frac{d}{dt}[\text{ln}(t)] \]
03

Differentiate the Outer Function

Differentiate the outer function \(\text{cos}(u)\) with respect to \(u\): \[ \frac{d}{du}[\text{cos}(u)] = -\text{sin}(u) \]
04

Differentiate the Inner Function

Differentiate the inner function \( \text{ln}(t) \) with respect to \( t \): \[ \frac{d}{dt}[\text{ln}(t)] = \frac{1}{t} \]
05

Combine the Derivatives

Combine the results from Steps 3 and 4 using the chain rule: \[ g'(t) = -\text{sin}(\text{ln}(t)) \times \frac{1}{t} \]
06

Simplify the Expression

Simplify the final expression to obtain the derivative: \[ g'(t) = -\frac{\text{sin}(\text{ln}(t))}{t} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Chain Rule
The chain rule is an essential differentiation technique when dealing with composite functions. It allows us to differentiate the outer function first and then multiply it by the derivative of the inner function. This rule is crucial when we have functions nested within each other, like in our example where we have a cosine function containing a logarithmic function.
The chain rule formula is usually articulated as:
\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]
By applying this rule, you can tackle complex differentiation problems with ease. More precisely, it enables step-by-step processing of each function layer. Make sure to correctly identify the outer and inner functions.
In our problem, we differentiate \[ g(t) = \cos(\ln(t)) \] by breaking it down into its components.
Differentiation Techniques
Differentiating complex functions often requires a good grasp of several techniques. Here, we focus on the necessity to handle both elementary and composite functions. Common techniques include:
  • Basic rule for polynomials: Power rule
  • Product rule for products of functions: Chain Rule
  • Quotient rule for ratios of functions: Quotient Rule
  • Trigonometric functions: Sine and Cosine rules
  • Logarithmic functions: Logarithm rule
In our example, we combined trigonometric differentiation with natural logarithms. Differentiating the outer cosine function results in:
\[ \frac{d}{du}[\cos(u)] = -\sin(u) \]
Then, differentiate the inner logarithmic function:
\[ \frac{d}{dt}[\ln(t)] = \frac{1}{t} \]
Finally, apply the chain rule to combine these derivatives.
Composite Functions
Composite functions are formed when one function is nested inside another. Understanding them is vital for effectively applying the chain rule.
In the problem given, we identify the composite function: \[ g(t) = \cos(\ln(t)) \]
Here, \[ \cos(u) \] is the outer function and \[ \ln(t) \] is the inner function. To differentiate composite functions, we need to:
  • Identify the outer and inner functions accurately.
  • Differentiate the outer function with respect to the inner variable.
  • Differentiate the inner function with respect to the main variable.
  • Multiply both derivatives according to the chain rule.
So, integrating these steps for \[ g(t) = \cos(\ln(t)) \], we get:
\[ g'(t) = \frac{d}{du}[\cos(u)] \cdot \frac{d}{dt}[\ln(t)] = -\sin(\ln(t)) \cdot \frac{1}{t} = -\frac{\sin(\ln(t))}{t} \]

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Most popular questions from this chapter

Find \(d y / d x\) in terms of \(x\) and \(y\) if \(x^{5} y-x-9 y-8=0\). \(\frac{d y}{d x}=\) _________

Find the derivative of the function \(h(r)\), below. It may be to your advantage to simplify first. $$ h(r)=\frac{r^{2}}{15 r+11} $$ \(h^{\prime}(r)=\) _______

Given \(F(2)=3, F^{\prime}(2)=4, F(4)=1, F^{\prime}(4)=5\) and \(G(1)=3, G^{\prime}(1)=4, G(4)=2, G^{\prime}(4)=7\) find each of the following. (Enter dne for any derivative that cannot be computed from this information alone.) A. \(H(4)\) if \(H(x)=F(G(x))\) _________ B. \(H^{\prime}(4)\) if \(H(x)=F(G(x))\) __________ C. \(H(4)\) if \(H(x)=G(F(x))\) ________ D. \(H^{\prime}(4)\) if \(H(x)=G(F(x))\) ________ E. \(H^{\prime}(4)\) if \(H(x)=F(x) / G(x)\) ___________

If a spherical tank of radius 4 feet has \(h\) feet of water present in the tank, then the volume of water in the tank is given by the formula $$ V=\frac{\pi}{3} h^{2}(12-h), $$ a. At what instantaneous rate is the volume of water in the tank changing with respect to the height of the water at the instant \(h=1 ?\) What are the units on this quantity? b. Now suppose that the height of water in the tank is being regulated by an inflow and outflow (e.g., a faucet and a drain) so that the height of the water at time \(t\) is given by the rule \(h(t)=\sin (\pi t)+1,\) where \(t\) is measured in hours \((\) and \(h\) is still measured in feet). At what rate is the height of the water changing with respect to time at the instant \(t=2 ?\) c. Continuing under the assumptions in (b), at what instantaneous rate is the volume of water in the tank changing with respect to time at the instant \(t=2 ?\) d. What are the main differences between the rates found in (a) and (c)? Include a discussion of the relevant units.

Find the slope of the tangent to the curve \(x^{3}+2 x y+y^{2}=64\) at (1,7). The slope is \(\square\). (Enter undef if the slope is not defined at this point.)

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