这是一份Bath巴斯大学MA20223作业代写的成功案

If we assume that $y(x)$ is a differentiable function of $x$, we can differentiate both sides of the equation
$$
x \cos (y(x))=e^{y(x)}
$$
and maintain the equality. Using the chain rule, this gives
$$
\cos (y(x))-x \sin (y(x)) y^{\prime}(x)=e^{y(x)} y^{\prime}(x) .
$$
Solving for $y^{\prime}(x)$ yields
$$
y^{\prime}(x)=\frac{\cos (y)}{e^{y}+x \sin (y)} .
$$
Substituting in $x=1$ and $y=0$ into the equation gives
$$
y^{\prime}(0)=1 .
$$

MA20223 COURSE NOTES ：

We can also use implicit differentiation to find the derivatives of rational powers of $x$. Finding derivatives of arbitrary powers of $x$ involves using the natural exponential and is left for later in this book.

Consider the function $y(x)=x^{\frac{n}{m}}$ where $n$ and $m$ are integers. Then $y^{m}=\left(x^{\frac{n}{m}}\right)^{m}=x^{n}$. Taking derivatives of both sides gives
$$
m y^{m-1} y^{\prime}(x)=n x^{n-1}
$$
or, since $y(x)=x^{n / m}$
$$
y^{\prime}(x)=\frac{n}{m} x^{n-1} x^{-(m-1) \frac{n}{m}}
$$
Simplifying the exponent of $x$ on the right side gives the desired result,
$$
\frac{d}{d x} x^{\frac{n}{m}}=\frac{n}{m} x^{\frac{n}{m}-1}
$$