Question 1

Y= 5x*4 tan X
Find dy/dx

Question 2

$\:\:\:\:\:\:\star\longrightarrow \sf \underline{y = 5\:x^4\:tanx}\\$

Differentiating with respect to x-

$\:\:\:\:\:\:\longrightarrow \sf(d)/(dx)\:y = (d)/(dx)\: 5\:x^4\:tanx\\$

$\:\:\:\:\:\:\longrightarrow \sf(dy)/(dx)\:= 5\:(d)/(dx)\: \:x^4\:tanx\\$

$\:\:\:\:\:\:\longrightarrow \sf(d)/(dx)\:y =5\:\bigg[tanx* (d)/(dx)\: \:x^4\:+x^4* (d)/(dx) tanx\bigg]\\$

$\pink{\star}\:\boxed{\sf\sf(d)/(dx)\bigg[f(x)\:g(x)\bigg] = f(x) (d)/(dx) g(x) + g(x) (d)/(dx) f(x)}\\$

$\:\:\:\:\:\:\longrightarrow \sf(dy)/(dx)= 5\:\bigg[tanx* 4\:x^3+ x^4 * sec^2x \bigg]\\$

$\qquad\pink{\star}\:\:\:\:\boxed{\sf (d)/(dx)x^n=nx^(n-1)}\\$

$\qquad\pink{\star}\:\:\:\:\boxed{\sf (d)/(dx)tanx=sec^2x}\\$

$\:\:\:\:\:\:\longrightarrow \sf(dy)/(dx)= 5\:x^3\:\bigg[4tanx+ x sec^2x \bigg]\\$

$\:\:\:\:\:\:\longrightarrow \sf\underline{(dy)/(dx)= \boxed{\sf5\:x^3\:\bigg[4tanx+ x sec^2x \bigg]}}\\$

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