Help me with my Calculus 1 example. Here is a photo

Ayanda Mazibuko

Answered question

2022-07-20

nick1337

Expert2023-05-28Added 777 answers

To determine the first-order partial derivative of the function

z = \ln (x + t^{2})

, we need to find the partial derivatives with respect to each variable, assuming that

t

is independent and

x

is the dependent variable.
To find

\frac{\partial z}{\partial x}

, we differentiate

z

with respect to

x

while treating

t

as a constant. The derivative of the natural logarithm function is given by

\frac{d}{d x} \ln (u) = \frac{1}{u} \cdot \frac{d u}{d x}

. Applying this rule, we have:

\frac{\partial z}{\partial x} = \frac{1}{x + t^{2}} \cdot \frac{d}{d x} (x + t^{2})

Now, differentiating

x + t^{2}

with respect to

x

, we get:

\frac{\partial z}{\partial x} = \frac{1}{x + t^{2}} \cdot (1 + 0) = \frac{1}{x + t^{2}}

Hence,

\frac{\partial z}{\partial x} = \frac{1}{x + t^{2}}

.
To find

\frac{\partial z}{\partial t}

, we differentiate

z

with respect to

t

while treating

x

as a constant. Applying the same rule as before, we have:

\frac{\partial z}{\partial t} = \frac{1}{x + t^{2}} \cdot \frac{d}{d t} (x + t^{2})

Differentiating

x + t^{2}

with respect to

t

, we get:

\frac{\partial z}{\partial t} = \frac{1}{x + t^{2}} \cdot (0 + 2 t) = \frac{2 t}{x + t^{2}}

Therefore,

\frac{\partial z}{\partial t} = \frac{2 t}{x + t^{2}}

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