Find the derivative of the given function.h(x)=4(x+3x)54h\left(x\right)=4^{\left(\sqrt{x}+3x\right)^{\frac54}}

54⋅4(x+3x)54⋅ln⁡4⋅(x+3x)14⋅(12x+3)\frac54\$$cdot4^{\left(\sqrt{x}+3x$$\$$right)^{\frac54}$$}\cdot\ln4\cdot\left(\sqrt{x}+3x\$$right)^{\frac14}$$\cdot\left(\frac{1}{2\sqrt{x}}+3\right)

Find the derivative of the given function. h ( x ) = 4 ( x + 3 x ) 5 4 h\left(x\right)=4^{\left(\sqrt{x}+3x\right)^{\frac54}}

Here, we wanna find the derivative of the function h of x equals 4 to the power of the square root of x plus 3 x, and everything in that exponent is raised to the power of 5 4ths. Now in order to find this derivative, we're going to have to apply the chain rule multiple times. So let's go ahead and get started. Now we wanna find our derivative here, h prime of x. So starting with our outermost function, we have this exponential function, 4 raised to the power of all of this stuff. Then as we work our way inside, in that exponent, we have all of this stuff being raised to the power of 5 4ths. Then inside of those parentheses, this is our innermost function, we have the square root of x plus 3 x. So we're starting from the outside, working our way in using the chain rule. So, again, that outermost function, 4 raised to the power of all of this stuff. I know that the derivative of an exponential function is going to be that base raised to the power of everything that is there. So here it is the square root of x plus 3x raised to the power of 5 4ths times the natural log of that base, in this case, 4. Now working our way inside here, we now want to find the derivative of the stuff raised to the power of 5 4ths, and we're multiplying by it. This is what our chain rule tells us. So we're going to apply the power rule here. We're going to pull that 5 4ths out in front of those parentheses. So that is 5 4ths times root x plus 3x, and that gets raised to the power of 5 4ths minus 1 using that Power Rule. 5 4th minus 1 gives us 1 fourth. Then finally, one more function working our way inside. We wanna multiply by the derivative of root x plus 3x. So one final derivative to take here, the square root of x. Remember that you can think of this as being x to the power of 1 half. So its derivative is 1 half x to the power of negative one half using the power rule. Then the derivative of 3x is just 3. Now we can rewrite this slightly. There is quite a bit going on. Typically, we want to have any constants out front. So I'm going to go ahead and pull that 5 fourths out front. So this is 5 over 4 times 4 to the power of the square root of x plus 3x raised to the power of 5 4ths times the natural log of 4. And then this function here, so times the square root of x plus 3x raised to the power of 1 4th. I'm just going to keep that as a fraction for now in that power. And then multiply it here by 1 over 2 root x plus 3. I'm just going to keep this factored as is. And this is my as is, and this is my full derivative here, h prime of x, having used the chain rule to work with all of these functions that are all composed with each other. If you have any questions here, feel free to let us know. I know that these chain rule problems can sometimes get a bit tricky. I'll see you in the next one.

Find the derivative of the given function.h(x)=4(x+3x)54h\left(x\right)=$$4^{\left(\sqrt{x}+3x$$\$$right)^{\frac54}$$}

54⋅4(x+3x)54⋅ln⁡4⋅(x+3x)14⋅(12x+3)\frac54\$$cdot4^{\left(\sqrt{x}+3x$$\$$right)^{\frac54}$$}\cdot\ln4\cdot\left(\sqrt{x}+3x\$$right)^{\frac14}$$\cdot\left(\frac{1}{2\sqrt{x}}+3\right)

54(x+3x)(12x+3)⋅4(x+3x)54\frac54\left(\sqrt{x}+3x\right)\left(\frac{1}{2\sqrt{x}}+3\right)\$$cdot4^{\left(\sqrt{x}+3x$$\$$right)^{\frac54}$$}

5ln⁡(4)4(x+3x)⋅4(x+3x)54\frac{5\ln\left(4\right)}{4}\left(\sqrt{x}+3x\right)\$$cdot4^{\left(\sqrt{x}+3x$$\$$right)^{\frac54}$$}

(x+3x)54⋅4(x+3x)54\left(\sqrt{x}+3x\$$right)^{\frac54}$$\$$cdot4^{\left(\sqrt{x}+3x$$\$$right)^{\frac54}$$}

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0. Functions7h 55m

Introduction to Functions
18m

Piecewise Functions
10m

Properties of Functions
9m

Common Functions
1h 8m

Transformations
5m

Combining Functions
27m

Exponent rules
32m

Exponential Functions
28m

Logarithmic Functions
24m

Properties of Logarithms
36m

Exponential & Logarithmic Equations
35m

Introduction to Trigonometric Functions
38m

Graphs of Trigonometric Functions
44m

Trigonometric Identities
47m

Inverse Trigonometric Functions
48m

1. Limits and Continuity2h 2m

Introduction to Limits
41m

Finding Limits Algebraically
40m

Continuity
40m

2. Intro to Derivatives1h 33m

Tangent Lines and Derivatives
30m

Derivatives as Functions
14m

Basic Graphing of the Derivative
23m

Differentiability
26m

3. Techniques of Differentiation3h 18m

Basic Rules of Differentiation
39m

Higher Order Derivatives
12m

Product and Quotient Rules
55m

Derivatives of Trig Functions
40m

The Chain Rule
49m

4. Applications of Derivatives2h 38m

Motion Analysis
36m

Implicit Differentiation
27m

Related Rates
59m

Linearization
13m

Differentials
21m

5. Graphical Applications of Derivatives6h 2m

Intro to Extrema
23m

Finding Global Extrema
50m

The First Derivative Test
1h 5m

Concavity
1h 1m

The Second Derivative Test
31m

Curve Sketching
44m

Applied Optimization
1h 25m

6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m

Derivatives of Exponential & Logarithmic Functions
1h 52m

Logarithmic Differentiation
20m

Derivatives of Inverse Trigonometric Functions
24m

7. Antiderivatives & Indefinite Integrals1h 26m

Antiderivatives
17m

Indefinite Integrals
25m

Integrals of Trig Functions
15m

Initial Value Problems
27m

8. Definite Integrals4h 44m

Estimating Area with Finite Sums
42m

Riemann Sums
1h 21m

Introduction to Definite Integrals
22m

Fundamental Theorem of Calculus
37m

Average Value of a Function
21m

Substitution
1h 19m

9. Graphical Applications of Integrals2h 27m

Area Between Curves
1h 18m

Introduction to Volume & Disk Method
1h 8m

10. Physics Applications of Integrals 3h 16m

Kinematics
1h 13m

Work
2h 3m

11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 31m

Integrals of Exponential Functions
40m

Integrals Involving Logarithmic Functions
58m

Integrals Involving Inverse Trigonometric Functions
52m

12. Techniques of Integration7h 41m

Integration by Parts
2h 32m

Partial Fractions
4h 29m

Improper Integrals
39m

13. Intro to Differential Equations2h 55m

Basics of Differential Equations
48m

Slope Fields
19m

Euler's Method
22m

Separable Differential Equations
1h 25m

14. Sequences & Series5h 36m

Sequences
1h 22m

Review of Factorials
11m

Series
44m

Convergence Tests
3h 17m

15. Power Series2h 19m

Introduction to Power Series
1h 9m

Taylor Series & Taylor Polynomials
1h 10m

16. Parametric Equations & Polar Coordinates7h 58m

Parametric Equations
1h 6m

Calculus with Parametric Curves
1h 13m

Polar Coordinates
2h 5m

Calculus in Polar Coordinates
55m

Conic Sections
2h 36m

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

Calculus

6. Derivatives of Inverse, Exponential, & Logarithmic Functions

Derivatives of Exponential & Logarithmic Functions

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Find the derivative of the given function.
$h (x) = 4^{{(\sqrt{x} + 3 x)}^{\frac{5}{4}}}$

$\frac{5}{4} (\sqrt{x} + 3 x) (\frac{1}{2 \sqrt{x}} + 3) \cdot 4^{{(\sqrt{x} + 3 x)}^{\frac{5}{4}}}$

$\frac{5 \ln (4)}{4} (\sqrt{x} + 3 x) \cdot 4^{{(\sqrt{x} + 3 x)}^{\frac{5}{4}}}$

$\frac{5}{4} \cdot 4^{{(\sqrt{x} + 3 x)}^{\frac{5}{4}}} \cdot \ln 4 \cdot {(\sqrt{x} + 3 x)}^{\frac{1}{4}} \cdot (\frac{1}{2 \sqrt{x}} + 3)$

${(\sqrt{x} + 3 x)}^{\frac{5}{4}} \cdot 4^{{(\sqrt{x} + 3 x)}^{\frac{5}{4}}}$

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Problema precedenteProssimo problema

Guida verificata passo dopo passo

Identify the function to differentiate: $ h(x) = 4(\sqrt{x} + 3x)^{\frac{5}{4}} $. This is a composite function, so we will use the chain rule.

Apply the chain rule: If $ y = (u(x))^n $, then $ \frac{dy}{dx} = n(u(x))^{n-1} \cdot \frac{du}{dx} $. Here, $ u(x) = \sqrt{x} + 3x $ and $ n = \frac{5}{4} $.

Differentiate $ u(x) = \sqrt{x} + 3x $: $ \frac{du}{dx} = \frac{1}{2\sqrt{x}} + 3 $.

Substitute back into the chain rule: $ \frac{dh}{dx} = 4 \cdot \frac{5}{4} (\sqrt{x} + 3x)^{\frac{1}{4}} \cdot (\frac{1}{2\sqrt{x}} + 3) $.

Simplify the expression: $ \frac{dh}{dx} = 5 (\sqrt{x} + 3x)^{\frac{1}{4}} \cdot (\frac{1}{2\sqrt{x}} + 3) $.