To find: Differentiation of (x³ cos x – 2^x tan x)

Formula used: (i) (uv)′ = u′v + uv′ (Leibnitz or product rule)

(ii)

(iii)

(iv)

(v)

Here we have two function (x³ cos x) and (2^x tan x)

We have two differentiate them separately

Let us assume g(x) = (x³ cos x)

And h(x) = (2^x tan x)

Therefore, f(x) = g(x) – h(x)

⇒ f’(x) = g’(x) – h’(x) … (i)

Applying product rule on g(x)

Let us take u = x³ and v = cos x

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

[x³ cosx]’ = 3x² × cosx + x³ × -sinx

= 3x²cosx - x³sinx

= x² (3cosx – x sinx)

g’(x) = x² (3cosx – x sinx)

Applying product rule on h(x)

Let us take u = 2^x and v = tan x

Putting the above obtained values in the formula:-

(uv)′ = u′v + uv′

[2^x tan x]’ = 2^x log2× tanx + 2^x × sec²x

= 2^x (log2tanx + sec²x)

h’(x) = 2^x (log2tanx + sec²x)

Putting the above obtained values in eqn. (i)

f’(x) = x² (3cosx – x sinx) - 2^x (log2tanx + sec²x)

Ans) x² (3cosx – x sinx) - 2^x (log2tanx + sec²x)