Question

y = 4^log2 (sin x) + 9^log3 (cos x) , then (log 2)(log 3) dy dx =​

Answer 1

Derivatives

Formula:

$\quad\mathsf{\frac{d}{dx}(a^{mx})=ma^{mx}.lna,\:(a>0)\:m\in\mathbb{R}}$

$\quad\mathsf{\frac{d}{dx}\{ln(sinx)\}=cotx}$

$\quad\mathsf{\frac{d}{dx}\{ln(cosx)\}=-tanx}$

Solution.

Let, $\mathsf{p(x)=4^{log_{2}(sinx)}}$

and $\mathsf{q(x)=9^{log_{3}(cosx)}}$

We do term by term derivative.

For $\mathsf{p(x)}$:

Differentiating with respect to $\mathsf{x}$, we get

$\quad \mathsf{\frac{d}{dx}\{p(x)\}=\frac{d}{dx}\{4^{log_{2}(sinx)}\}}$

$\Rightarrow \mathsf{p'(x)=4^{log_{2}(sinx)}.ln4.\frac{d}{dx}\{log_{2}(sinx)\}}$

$\Rightarrow \mathsf{p'(x)=4^{log_{2}(sinx)}.ln4.\frac{d}{dx}\left(\frac{ln(sinx)}{ln2}\right)}$

$\Rightarrow \mathsf{p'(x)=4^{log_{2}(sinx)}.\frac{ln4}{ln2}.cotx}$

For $\mathsf{q(x)}$:

Differentiating with respect to $\mathsf{x}$, we get

$\quad \mathsf{\frac{d}{dx}\{q(x)\}=\frac{d}{dx}\{9^{log_{3}(cosx)}\}}$

$\Rightarrow \mathsf{q'(x)=9^{log_{3}(cosx)}.ln9.\frac{d}{dx}\{log_{3}(cosx)\}}$

$\Rightarrow \mathsf{q'(x)=9^{log_{3}(cosx)}.ln9.\frac{d}{dx}\left(\frac{ln(cosx)}{ln3}\right)}$

$\Rightarrow \mathsf{q'(x)=9^{log_{3}(cosx)}.\frac{ln9}{ln3}.(-tanx)}$

Now, $\mathsf{\frac{dy}{dx}=p'(x)+q'(x)}$

$\Rightarrow \mathsf{\frac{dy}{dx}=4^{log_{2}(sinx)}.\frac{ln4}{ln2}.cotx-9^{log_{3}(cosx)}.\frac{ln9}{ln3}.tanx}$

$\Rightarrow \mathsf{ln2.ln3.\frac{dy}{dx}=4^{log_{2}(sinx)}.ln3.ln4.cotx-9^{log_{3}(cosx)}.ln2.ln9.tanx}$

$\Rightarrow \boxed{\color{blue}\mathsf{ln2.ln3.\frac{dy}{dx}=2.ln3 .ln2.[4^{log_{2}(sinx)}.cotx-9^{log_{3}(cosx)}.tanx]}}$