# A complicated derivative using chain rule

In summary, using the chain rule, the derivative dz/dt of the function z = x^{2}y^{3} + e^{y}\cos x can be found without substitution to be 2e^xy(2x + x^2 .y)(t) - (x^3 .e^xy)(-t^-2) + 3e^xy(x^2 .y^2)(4cos(4t)).

## Homework Statement

I have a function z, and I need to find the derivative dz/dt "using the chain rule without substitution"

## Homework Equations

$$z = x^{2}y^{3} + e^{y}\cos x$$
$$x = \log(t^{2})$$
$$y = \sin(4t)$$

## The Attempt at a Solution

$$\frac{\mathrm{d} z}{\mathrm{d} t} = 2xy^{3}+3x^{2}y^{2}+e^{y}\cos x-e^{y}\sin x$$
$$\frac{\mathrm{d} z}{\mathrm{d} t} = 2\log(t^{2})\sin(4t)^{3}+3(\log(t^{2}))^{2}(\sin(4t))^{2}+e^{\sin(4t)}(\cos (\log(t^{2}))-\sin (\log(t^{2})))$$
$$\frac{\mathrm{d} z}{\mathrm{d} t} = 4\log(t)\sin(4t)^{3}+3(2\log(t))^{2}(sin(4t))^{2}+e^{sin(4t)}(\cos (2\log(t))-sin (2\log(t)))$$

I'm a bit stuck at this point, I feel like I should be able to do something with the log^2 and the e^sin . (cos(log) - sin(log))?

Last edited:
That's not right.. Now you have x=x(t), y=y(t), z=z(t) so you'd get
$$\frac{dz}{dt} = 3 x^2 y^2 \frac{dy}{dt} + ...$$
etc. This is the chain rule, and it's the most important differentiation rule, so you should definitely read on it further :-)

I just wanted to say thanks, I studied my book, and I did approach this in the wrong way! I won't do the latex script, but my answer is (without substitution):

dz/dt = ∂z/∂x . dx/dt + ∂z/∂y . dy/dt

= 2e^xy .(2x + x^2 .y)(t) - (x^3 .e^xy)(-t^-2)

## 1. What is the chain rule?

The chain rule is a mathematical concept used in calculus to help calculate the derivative of a function composed of two or more functions. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function. In other words, it helps us find the rate of change of a function within a function.

## 2. How do you use the chain rule?

To use the chain rule, you first identify the outer function and the inner function. Then, you take the derivative of the outer function and multiply it by the derivative of the inner function. If there are multiple functions nested within each other, you continue this process until you reach the innermost function. Finally, you multiply all the derivatives together to find the overall derivative of the composite function.

## 3. Why is the chain rule important?

The chain rule is important because it allows us to find the derivative of complex functions that cannot be easily calculated using basic derivative rules. It is a fundamental concept in calculus and is used in many real-world applications, such as physics, economics, and engineering.

## 4. Can you provide an example of using the chain rule?

Sure, let's say we have the function y = (2x + 1)^3. To find the derivative of this function, we first identify the outer function as (2x + 1)^3 and the inner function as 2x + 1. The derivative of the outer function is 3(2x + 1)^2 and the derivative of the inner function is 2. Therefore, the overall derivative is 3(2x + 1)^2 * 2 = 6(2x + 1)^2.

## 5. Are there any common mistakes when using the chain rule?

Yes, some common mistakes when using the chain rule include forgetting to take the derivative of the outer function, not properly identifying the inner and outer functions, and not multiplying all the derivatives together. It is important to pay attention to the order of operations and carefully follow the steps of the chain rule to avoid these mistakes.

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