Animated Solution for Mathematics - Differentiation: Let $y=e^{x\sin x^3}+(\tan x{)}^x$. Find $\frac{dy}{dx}$.

Taming the Calculus Beast

Calculus is often perceived as a battlefield of complex expressions, but at its heart, it is an art of decomposition. When you look at a function like $y=e^{x\sin x^3}+(\tan x{)}^x$, it is natural to feel a moment of hesitation.

It looks like a monster. But remember, even the most complex structures are built from simple, manageable blocks. Our strategy today is simple: divide and conquer.

The Strategy of Separation

We begin by defining our two components: $u=e^{x\sin x^3}$ and $v=(\tan x{)}^x$.

By treating $y$ as $y=u+v$, we can invoke the sum rule of differentiation:

This simple step transforms one overwhelming problem into two distinct, focused tasks. We are no longer fighting a monster; we are solving two separate puzzles.

Unpacking the First Beast ($u$)

Let us tackle $u=e^{x\sin x^3}$. This is a classic application of the chain rule. We know that the derivative of $e^f$ is $e^f\cdot f^{\prime }$.

So, our first step is:

Now, the focus shifts to the exponent: $x\sin x^3$. This is a product of two functions, $x$ and $\sin x^3$. Here, we apply the product rule: $f\cdot g^{\prime }+g\cdot f^{\prime }$.

This gives us:

To differentiate $\sin x^3$, we need the chain rule again. The derivative of $\sin$ is $\cos$, and the derivative of $x^3$ is $3x^2$. Thus, $\frac{d}{dx}(\sin x^3)=3x^2\cos x^3$.

Putting it all together, we get:

The Logarithmic Equalizer ($v$)

Now, we turn to $v=(\tan x{)}^x$. As we discussed, the variable exponent is a trap for the unwary. We use logarithmic differentiation to bring that exponent down to earth.

Taking the natural log of both sides, we get $\ln v=x\ln (\tan x)$. Now, we differentiate both sides with respect to $x$.

On the left, we have $\frac{1}{v}\frac{dv}{dx}$. On the right, we use the product rule again:

For the term $\frac{d}{dx}(\ln \tan x)$, we apply the chain rule: $\frac{1}{\tan x}\cdot {\sec }^{2}x$. This simplifies beautifully.

Since $\frac{1}{\tan x}=\frac{\cos x}{\sin x}$ and ${\sec }^{2}x=\frac{1}{{\cos }^{2}x}$, the product becomes $\frac{1}{\sin x\cos x}$. By multiplying the numerator and denominator by $2$, we get $\frac{2}{\sin 2x}$.

The Final Synthesis

We are almost there. We have:

Multiplying by $v$ gives us:

Finally, we combine our two results to find the derivative of the original function:

Look at that result. It is elegant, precise, and entirely derived from the systematic application of fundamental rules. You didn't just solve a problem; you navigated a complex landscape of calculus. Keep this systematic mindset, and no expression will ever be too large for you to handle.